NCERT Solutions: Conic Sections

# Conic Sections NCERT Solutions | Mathematics (Maths) Class 11 - Commerce

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NCERT QUESTION - (Conic Sections)

Question 1: Find the equation of the circle with centre (0, 2) and radius 2

ANSWER : - The equation of a circle with centre (hk) and radius is given as

(x­ – h)2 + (y ­– k)2r2

It is given that centre (hk) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x – 0)2 + (y – 2)2 = 22

x2y2 + 4 ­– 4 +y = 4

x2  + y2 ­– 4y = 0

Question 2: Find the equation of the circle with centre (–2, 3) and radius 4

ANSWER : - The equation of a circle with centre (hk) and radius is given as

(x­ – h)2 + (y ­– k)2r2

It is given that centre (hk) = (–2, 3) and radius (r) = 4.

Therefore, the equation of the circle is

(x + 2)2 + (y – 3)2 = (4)2

x2 + 4x + 4 + y2 – 6y + 9 = 16

x2y2+ 4– 6y – 3 = 0

Question 3: Find the equation of the circle with centre    and radius

ANSWER : - The equation of a circle with centre (hk) and radius is given as

(x­ – h)2 + (y ­– k)2r2

It is given that centre (hk) = and radius (r) = .

Therefore, the equation of the circle is

Question 4: Find the equation of the circle with centre (1, 1) and radius

ANSWER : - The equation of a circle with centre (hk) and radius is given as

(x­ – h)2 + (y ­– k)2r2

It is given that centre (hk) = (1, 1) and radius (r) =  .

Therefore, the equation of the circle is

Question 5: Find the equation of the circle with centre (–a, –b) and radius

ANSWER : - The equation of a circle with centre (hk) and radius is given as

(x­ – h)2 + (y ­– k)2r2

It is given that centre (hk) = (–a, –b) and radius (r) = .

Therefore, the equation of the circle is

Question 6: Find the centre and radius of the circle (x + 5)2 + (y – 3)2 = 36

ANSWER : - The equation of the given circle is (x + 5)2 + (y – 3)2 = 36.

(x + 5)2 + (y – 3)2 = 36

⇒ {x – (–5)}2 + (y – 3)2 = 62, which is of the form (x – h)2 + (y – k)2r2, where h = –5, k = 3, and r = 6.

Thus, the centre of the given circle is (–5, 3), while its radius is 6.

Question 7: Find the centre and radius of the circle x2y2 – 4x – 8y – 45 = 0

ANSWER : - The equation of the given circle is x2y2 – 4x – 8y – 45 = 0.

x2  + y2 – 4x – 8y – 45 = 0

⇒ (x2 – 4x)+ (y– 8y) = 45

⇒ {x2 – 2(x)(2) + 22}+ {y2 – 2(y)(4)+ 42} – 4 –16 = 45

⇒ (x – 2)2 + (y –4)2 = 65

⇒ (x – 2)2 + (y –4)2 =  , which is of the form (x – h)2 + (y – k)2r2, where h = 2, k = 4, and  .

Thus, the centre of the given circle is (2, 4), while its radius is  .

Question 8: Find the centre and radius of the circle x2y2 – 8x+  10y – 12 = 0

ANSWER : - The equation of the given circle is x2y2 – 8x + 10y – 12 = 0.

x2  + y2 – 8x  + 10y – 12 = 0

⇒ (x2 – 8x) + (y 10y) = 12

⇒ {x2 – 2(x)(4) +  42}  + {y 2(y)(5)  + 52}– 16 – 25 = 12

⇒ (x – 4)2 + (y + 5)2 = 53

which is of the form (x – h)2 + (y – k)2r2, where h = 4, k = –5, and  .

Thus, the centre of the given circle is (4, –5), while its radius is  .

Question 9: Find the centre and radius of the circle 2x2 + 2y2 – x = 0

ANSWER : - The equation of the given circle is 2x2+  2y2 – x = 0.

which is of the form (x – h)2 + (y – k)2r2, where h , k = 0, and  .

Thus, the centre of the given circle is , while its radius is .

Question 10: Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4xy = 16.

ANSWER : - Let the equation of the required circle be (x – h)2 + (y – k)2r2.

Since the circle passes through points (4, 1) and (6, 5),

(4 – h)2 + (1 – k)2r2 … (1)

(6 – h)2 + (5 – k)2r2 … (2)

Since the centre (h, k) of the circle lies on line 4xy = 16,

4h  + k = 16 … (3)

From equations (1) and (2), we obtain

(4 – h)2 + (1 – k)= (6 – h)2 + (5 – k)2

⇒ 16 – 8h  + h2  1 – 2kk2 = 36 – 12hh2 + 25 – 10kk2

⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k

⇒ 4h + 8k = 44

⇒ h + 2k = 11 … (4)

On solving equations (3) and (4), we obtain h = 3 and k = 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)2 + (1 – 4)2r2

⇒ (1)2 + (– 3)2r2

⇒ 1+ 9 = r2

⇒ r2 = 10

⇒

Thus, the equation of the required circle is

(x – 3)2+  (y – 4)2

x2 – 6x + 9+  y2 ­– 8y + 16 = 10

x2y2 – 6x – 8y + 15 = 0

Question 11: Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line – 3y – 11 = 0.

ANSWER : - Let the equation of the required circle be (x – h)2 + (y – k)2r2.

Since the circle passes through points (2, 3) and (–1, 1),

(2 – h)2 + (3 – k)2r2 … (1)

(–1 – h)2 + (1 – k)2r2 … (2)

Since the centre (h, k) of the circle lies on line – 3y – 11 = 0,

– 3k = 11 … (3)

From equations (1) and (2), we obtain

(2 – h)2 + (3 – k)2 = (–1 – h)2 + (1 – k)2

⇒ 4 – 4hh2 + 9 – 6kk2 = 1 +2h  + h2  +1 – 2kk2

⇒ 4 – 4h + 9 – 6k = 1 +2h + 1 – 2k

⇒ 6h + 4k = 11 … (4)

On solving equations (3) and (4), we obtain .

On substituting the values of h and k in equation (1), we obtain

Thus, the equation of the required circle is

Question 12: Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

ANSWER : - Let the equation of the required circle be (x – h)2 + (y – k)2r2.

Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

Now, the equation of the circle becomes (x – h)2y2 = 25.

It is given that the circle passes through point (2, 3).

When h = –2, the equation of the circle becomes

(x + 2)2y2 = 25

x2 + 4x + 4 + y2 = 25

x2y2+  4x – 21 = 0

When h = 6, the equation of the circle becomes

(x – 6) + y2 = 25

x2 – 12x + 36+  y2 = 25

x2y2 – 12x + 11 = 0

Question 13: Find the equation of the circle passing through (0, 0) and making intercepts and b on the coordinate axes.

ANSWER : - Let the equation of the required circle be (x – h)2 + (y – k)2r2.

Since the centre of the circle passes through (0, 0),

(0 – h)2 + (0 – k)2r2

⇒ h2k2r2

The equation of the circle now becomes (x – h)2 + (y – k)2h2  + k2.

It is given that the circle makes intercepts and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,

(a – h)2 + (0 – k)2h2k2 … (1)

(0 – h)2 + (b – k)2h2k2 … (2)

From equation (1), we obtain

a2 – 2ah  + h2k2h2k2

⇒ a2 – 2ah = 0

⇒ a(a – 2h) = 0

⇒ a = 0 or (a – 2h) = 0

However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h = .

From equation (2), we obtain

h2b2 – 2bkk2h2k2

⇒ b2 – 2bk = 0

⇒ b(b – 2k) = 0

⇒ b = 0 or(b – 2k) = 0

However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k = .

Thus, the equation of the required circle is

Question 14: Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

ANSWER : - The centre of the circle is given as (hk) = (2, 2).

Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).

Thus, the equation of the circle is

Question 15: Does the point (–2.5, 3.5) lie inside, outside or on the circle x2y2 = 25?

ANSWER : - The equation of the given circle is x2y2 = 25.

x2y2 = 25

⇒ (– 0)2 + (y – 0)2 = 52, which is of the form (x – h)2 + (y – k)2r2, where h = 0, k = 0, and r = 5.

∴Centre = (0, 0) and radius = 5

Distance between point (–2.5, 3.5) and centre (0, 0)

Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.

Question 16: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 12x

ANSWER : - The given equation is y2 = 12x.

Here, the coefficient of is positive. Hence, the parabola opens towards the right.

On comparing this equation with y= 4ax, we obtain

4a = 12 ⇒ a = 3

∴Coordinates of the focus = (a, 0) = (3, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of direcctrix, x = –i.e., = – 3 i.e., x + 3 = 0

Length of latus rectum = 4a = 4 × 3 = 12

Question 17: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = 6y

ANSWER : - The given equation is x2 = 6y.

Here, the coefficient of is positive. Hence, the parabola opens upwards.

On comparing this equation with x2 = 4ay, we obtain

∴Coordinates of the focus = (0, a) =

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix,

Length of latus rectum = 4a = 6

Question 18: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = – 8x

ANSWER : - The given equation is y2 = –8x.

Here, the coefficient of is negative. Hence, the parabola opens towards the left.

On comparing this equation with y2 = –4ax, we obtain

–4a = –8 ⇒ a = 2

∴Coordinates of the focus = (–a, 0) = (–2, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of directrix, xa i.e., x = 2

Length of latus rectum = 4a = 8

Question 19: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = – 16y

ANSWER : - The given equation is x2 = –16y.

Here, the coefficient of is negative. Hence, the parabola opens downwards.

On comparing this equation with x2 = – 4ay, we obtain

–4a = –16 ⇒ a = 4

∴Coordinates of the focus = (0, –a) = (0, –4)

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix, ya i.e., y = 4

Length of latus rectum = 4a = 16

Question 20: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 10x

ANSWER : - The given equation is y2 = 10x.

Here, the coefficient of is positive. Hence, the parabola opens towards the right.

On comparing this equation with y= 4ax, we obtain

∴Coordinates of the focus = (a, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of directrix,

Length of latus rectum = 4a = 10

Question 21: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = –9y

ANSWER : - The given equation is x2 = –9y.

Here, the coefficient of is negative. Hence, the parabola opens downwards.

On comparing this equation with x2 = –4ay, we obtain

∴Coordinates of the focus =

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix,

Length of latus rectum = 4a = 9

Question 22: Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6

ANSWER : - Focus (6, 0); directrix, x = –6

Since the focus lies on the x-axis, the x-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form y2 = 4ax or

y2 = – 4ax.

It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis. Hence, the parabola is of the form y2 = 4ax.

Here, a = 6

Thus, the equation of the parabola is y2 = 24x.

Question 23: Find the equation of the parabola that satisfies the following conditions: Focus (0, –3); directrix y = 3

ANSWER : - Focus = (0, –3); directrix y = 3

Since the focus lies on the y-axis, the y-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form x2 = 4ay or

x= – 4ay.

It is also seen that the directrix, y = 3 is above the x-axis, while the focus

(0, –3) is below the x-axis. Hence, the parabola is of the form x2 = –4ay.

Here, a = 3

Thus, the equation of the parabola is x2 = –12y.

Question 24: Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0); focus (3, 0)

ANSWER : - Vertex (0, 0); focus (3, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y2 = 4ax.

Since the focus is (3, 0), a = 3.

Thus, the equation of the parabola is y2 = 4 × 3 × x, i.e., y2 = 12x

Question 25: Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (–2, 0)

ANSWER : - Vertex (0, 0) focus (–2, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y2 = –4ax.

Since the focus is (–2, 0), a = 2.

Thus, the equation of the parabola is y2 = –4(2)x, i.e., y2 = –8x

Question 26: Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis

ANSWER : - Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the form y2 = 4ax or y2 = –4ax.

The parabola passes through point (2, 3), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form y2 = 4ax, while point

(2, 3) must satisfy the equation y2 = 4ax.

Thus, the equation of the parabola is

Question 27: Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis

ANSWER : - Since the vertex is (0, 0) and the parabola is symmetric about the y-axis, the equation of the parabola is either of the form x2 = 4ay or x2 = –4ay.

The parabola passes through point (5, 2), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form x2 = 4ay, while point

(5, 2) must satisfy the equation x2 = 4ay.

Thus, the equation of the parabola is

Question 28: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

ANSWER : - The given equation is .

Here, the denominator of  is greater than the denominator of .

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with , we obtain = 6 and b = 4.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (6, 0) and (–6, 0).

Length of major axis = 2a = 12

Length of minor axis = 2b = 8

Length of latus rectum

Question 29: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

ANSWER : - The given equation is .

Here, the denominator of  is greater than the denominator of .

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with , we obtain = 2 and a = 5.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (0, 5) and (0, –5)

Length of major axis = 2a = 10

Length of minor axis = 2b = 4

Length of latus rectum

Question 30: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

ANSWER : - The given equation is .

Here, the denominator of  is greater than the denominator of .

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with , we obtain = 4 and b = 3.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are .

Length of major axis = 2a = 8

Length of minor axis = 2b = 6

Length of latus rectum

Question 31: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

ANSWER : - The given equation is .

Here, the denominator of  is greater than the denominator of .

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with , we obtain = 5 and a = 10.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (0, ±10).

Length of major axis = 2a = 20

Length of minor axis = 2b = 10

Length of latus rectum

Question 32: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

ANSWER : - The given equation is .

Here, the denominator of  is greater than the denominator of .

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with , we obtain = 7 and b = 6.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (± 7, 0).

Length of major axis = 2a = 14

Length of minor axis = 2b = 12

Length of latus rectum

Question 33: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

ANSWER : - The given equation is .

Here, the denominator of  is greater than the denominator of .

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with , we obtain = 10 and a = 20.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (0, ±20)

Length of major axis = 2a = 40

Length of minor axis = 2b = 20

Length of latus rectum

Question 34: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 36x2 + 4y2 = 144

ANSWER : - The given equation is 36x2 + 4y2 = 144.

It can be written as

Here, the denominator of  is greater than the denominator of .

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing equation (1) with , we obtain = 2 and a = 6.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (0, ±6).

Length of major axis = 2= 12

Length of minor axis = 2b = 4

Length of latus rectum

Question 35: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 16x2y2 = 16

ANSWER : - The given equation is 16x2y2 = 16.

It can be written as

Here, the denominator of  is greater than the denominator of .

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing equation (1) with , we obtain = 1 and a = 4.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (0, ±4).

Length of major axis = 2a = 8

Length of minor axis = 2b = 2

Length of latus rectum

Question 36: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 4x2 + 9y2 = 36

ANSWER : - The given equation is 4x2 + 9y2 = 36.

It can be written as

Here, the denominator of  is greater than the denominator of .

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with , we obtain = 3 and b = 2.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (±3, 0).

Length of major axis = 2a = 6

Length of minor axis = 2b = 4

Length of latus rectum

Question 37: Find the equation for the ellipse that satisfies the given conditions: Vertices (±5, 0), foci (±4, 0)

ANSWER : - Vertices (±5, 0), foci (±4, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form , where is the semi-major axis.

Accordingly, a = 5 and c = 4.

It is known that .

Thus, the equation of the ellipse is .

Question 38: Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ±13), foci (0, ±5)

ANSWER : - Vertices (0, ±13), foci (0, ±5)

Here, the vertices are on the y-axis.

Therefore, the equation of the ellipse will be of the form , where is the semi-major axis.

Accordingly, a = 13 and c = 5.

It is known that .

Thus, the equation of the ellipse is .

Question 39: Find the equation for the ellipse that satisfies the given conditions: Vertices (±6, 0), foci (±4, 0)

ANSWER : - Vertices (±6, 0), foci (±4, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form , where is the semi-major axis.

Accordingly, a = 6, c = 4.

It is known that .

Thus, the equation of the ellipse is .

Question 40: Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (±3, 0), ends of minor axis (0, ±2)

ANSWER : - Ends of major axis (±3, 0), ends of minor axis (0, ±2)

Here, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form , where is the semi-major axis.

Accordingly, a = 3 and b = 2.

Thus, the equation of the ellipse is .

Question 41: Find the equation for the ellipse that satisfies the given conditions: Ends of major axis , ends of minor axis (±1, 0)

ANSWER : - Ends of major axis , ends of minor axis (±1, 0)

Here, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form , where is the semi-major axis.

Accordingly, a =   and b = 1.

Thus, the equation of the ellipse is .

Question 42: Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (±5, 0)

ANSWER : - Length of major axis = 26; foci = (±5, 0).

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form , where is the semi-major axis.

Accordingly, 2a = 26 ⇒ a = 13 and c = 5.

It is known that .

Thus, the equation of the ellipse is .

Question 43: Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ±6)

ANSWER : - Length of minor axis = 16; foci = (0, ±6).

Since the foci are on the y-axis, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form , where is the semi-major axis.

Accordingly, 2b = 16 ⇒ b = 8 and c = 6.

It is known that .

Thus, the equation of the ellipse is .

Question 44: Find the equation for the ellipse that satisfies the given conditions: Foci (±3, 0), a = 4

ANSWER : - Foci (±3, 0), a = 4

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form , where is the semi-major axis.

Accordingly, c = 3 and a = 4.

It is known that .

Thus, the equation of the ellipse is .

Question 45: Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin; foci on the axis.

ANSWER : - It is given that b = 3, c = 4, centre at the origin; foci on the axis.

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form , where is the semi-major axis.

Accordingly, b = 3, c = 4.

It is known that .

Thus, the equation of the ellipse is .

Question 46: Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

ANSWER : - Since the centre is at (0, 0) and the major axis is on the y-axis, the equation of the ellipse will be of the form

The ellipse passes through points (3, 2) and (1, 6). Hence,

On solving equations (2) and (3), we obtain b2 = 10 and a2 = 40.

Thus, the equation of the ellipse is .

Question 47: Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

ANSWER : - Since the major axis is on the x-axis, the equation of the ellipse will be of the form

The ellipse passes through points (4, 3) and (6, 2). Hence,

On solving equations (2) and (3), we obtain a2 = 52 and b2 = 13.

Thus, the equation of the ellipse is .

Question 48: Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola

ANSWER : - The given equation is

On comparing this equation with the standard equation of hyperbola i.e., , we obtain a = 4 and b = 3.

We know that a2b2c2.

Therefore,

The coordinates of the foci are (±5, 0).

The coordinates of the vertices are (±4, 0).

Length of latus rectum

Question 49: Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola

ANSWER : - The given equation is .

On comparing this equation with the standard equation of hyperbola i.e., , we obtain a = 3 and .

We know that a2b2c2.

Therefore,

The coordinates of the foci are (0, ±6).

The coordinates of the vertices are (0, ±3).

Length of latus rectum

Question 50: Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y2 – 4x2 = 36

ANSWER : - The given equation is 9y2 – 4x2 = 36.

It can be written as

9y2 – 4x2 = 36

On comparing equation (1) with the standard equation of hyperbola i.e., , we obtain a = 2 and b = 3.

We know that a2b2c2.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are .

Length of latus rectum

Question 51: Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x2 – 9y2 = 576

ANSWER : - The given equation is 16x2 – 9y2 = 576.

It can be written as

16x2 – 9y2 = 576

On comparing equation (1) with the standard equation of hyperbola i.e., , we obtain a = 6 and b = 8.

We know that a2  + b2c2.

Therefore,

The coordinates of the foci are (±10, 0).

The coordinates of the vertices are (±6, 0).

Length of latus rectum

Question 52: Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y2 – 9x2 = 36

ANSWER : - The given equation is 5y2 – 9x2 = 36.

On comparing equation (1) with the standard equation of hyperbola i.e., , we obtain a and b = 2.

We know that a2  + b2c2.

Therefore, the coordinates of the foci are .

The coordinates of the vertices are .

Length of latus rectum

Question 53: Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y2 – 16x2 = 784

ANSWER : - The given equation is 49y2 – 16x2 = 784.

It can be written as
49y2 – 16x2 = 784

On comparing equation (1) with the standard equation of hyperbola i.e., , we obtain a = 4 and b = 7.

We know that a2  + b2c2.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (0, ±4).

Length of latus rectum

Question 54: Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0)

ANSWER : - Vertices (±2, 0), foci (±3, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form .

Since the vertices are (±2, 0), = 2.

Since the foci are (±3, 0), c = 3.

We know that a2  + b2c2.

Thus, the equation of the hyperbola is .

Question 55: Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci (0, ±8)

ANSWER : - Vertices (0, ±5), foci (0, ±8)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form

Since the vertices are (0, ±5), = 5.

Since the foci are (0, ±8), c = 8.

We know that a2  + b2c2.

Thus, the equation of the hyperbola is .

Question 56: Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci (0, ±5)

ANSWER : - Vertices (0, ±3), foci (0, ±5)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form .

Since the vertices are (0, ±3), = 3.

Since the foci are (0, ±5), c = 5.

We know that a2  + b2c2.

∴32b2 = 52

⇒ b2 = 25 – 9 = 16

Thus, the equation of the hyperbola is .

Question 57: Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the transverse axis is of length 8.

ANSWER : - Foci (±5, 0), the transverse axis is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci are (±5, 0), c = 5.

Since the length of the transverse axis is 8, 2a = 8 ⇒ a = 4.

We know that a2b2c2.

∴42b2 = 52

⇒ b2 = 25 – 16 = 9

Thus, the equation of the hyperbola is .

Question 58: Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13), the conjugate axis is of length 24.

ANSWER : - Foci (0, ±13), the conjugate axis is of length 24.

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci are (0, ±13), c = 13.

Since the length of the conjugate axis is 24, 2b = 24 ⇒ b = 12.

We know that a2  + b2c2.

a2 + 122 = 132

⇒ a2 = 169 – 144 = 25

Thus, the equation of the hyperbola is .

Question 59: Find the equation of the hyperbola satisfying the give conditions: Foci , the latus rectum is of length 8.

ANSWER : - Foci , the latus rectum is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci are c = .

Length of latus rectum = 8

We know that a2  + b2c2.

a2 + 4a = 45

⇒ a2 + 4a – 45 = 0

⇒ a2 + 9a – 5a – 45 = 0

⇒ (a + 9) (a – 5) = 0

⇒ a = –9, 5

Since a is non-negative, = 5.

b2 = 4= 4 × 5 = 20

Thus, the equation of the hyperbola is .

Question 60: Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12

ANSWER : - Foci (±4, 0), the latus rectum is of length 12.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci are (±4, 0), c = 4.

Length of latus rectum = 12

We know that a2  + b2c2.

a2 + 6a = 16

⇒ a2 + 6a – 16 = 0

⇒ a2 + 8a – 2a – 16 = 0

⇒ (a  8) (a – 2) = 0

⇒ a = –8, 2

Since a is non-negative, = 2.

b2 = 6= 6 × 2 = 12

Thus, the equation of the hyperbola is  .

Question 61: Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0),

ANSWER : - Vertices (±7, 0),

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form .

Since the vertices are (±7, 0), = 7.

It is given that

We know that a2b2c2.

Thus, the equation of the hyperbola is .

Question 62: Find the equation of the hyperbola satisfying the give conditions: Foci , passing through (2, 3)

ANSWER : - Foci , passing through (2, 3)

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci are c = .

We know that a2b2c2.

∴ a2b2 = 10

⇒ b2 = 10 – a2 … (1)

Since the hyperbola passes through point (2, 3),

From equations (1) and (2), we obtain

In hyperbola, ca, i.e., c2a2

∴ a2 = 5

⇒ b2 = 10 – a2 = 10 – 5 = 5

Thus, the equation of the hyperbola is .

Question 63: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

ANSWER : - The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form y2 = 4ax (as it is opening to the right).

Since the parabola passes through point A (10, 5), 102 = 4a(5)

⇒ 100 = 20a

Therefore, the focus of the parabola is (a, 0) = (5, 0), which is the mid-point of the diameter.

Hence, the focus of the reflector is at the mid-point of the diameter.

Question 64: An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

ANSWER : - The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the positive y-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).

It can be clearly seen that the parabola passes through point .

Therefore, the arch is in the form of a parabola whose equation is .

When y = 2 m,

Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23 m.

Question 65: The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

ANSWER : - The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y-axis. This can be diagrammatically represented as

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadway, 18 m from the middle.

Here, AB = 30 m, OC = 6 m, and .

The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).

The coordinates of point A are (50, 30 – 6) = (50, 24).

Since A (50, 24) is a point on the parabola,

∴Equation of the parabola,  or 6x2 = 625y

The x-coordinate of point D is 18.

Hence, at x = 18,

∴DE = 3.11 m

DF = DE EF = 3.11 m 6 m = 9.11 m

Thus, the length of the supporting wire attached to the roadway 18 m from the middle is approximately 9.11 m.

Question 66: An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

ANSWER : - Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2 m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis. Hence, the semi-ellipse can be diagrammatically represented as

The equation of the semi-ellipse will be of the form , where a is the semi-major axis

Accordingly, 2a = 8 ⇒ a = 4

b = 2

Therefore, the equation of the semi-ellipse is

Let A be a point on the major axis such that AB = 1.5 m.

Draw AC⊥ OB.

OA = (4 – 1.5) m = 2.5 m

The x-coordinate of point C is 2.5.

On substituting the value of x with 2.5 in equation (1), we obtain

∴AC = 1.56 m

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

Question 67: A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

ANSWER : - Let AB be the rod making an angle θ with OX and P (x, y) be the point on it such that AP = 3 cm.

Then, PB = AB – AP = (12 – 3) cm = 9 cm [AB = 12 cm]

From P, draw PQ⊥OY and PR⊥OX.

In ΔPBQ,

In ΔPRA,

Thus, the equation of the locus of point P on the rod is .

Question 68: Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.

ANSWER : - The given parabola is x2 = 12y.

On comparing this equation with x2 = 4ay, we obtain 4a = 12 ⇒ a = 3

∴The coordinates of foci are S (0, a) = S (0, 3)

Let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At y = 3, x2 = 12 (3) ⇒ x2 = 36 ⇒ x = ±6

∴The coordinates of A are (–6, 3), while the coordinates of B are (6, 3).

Therefore, the vertices of ΔOAB are O (0, 0), A (–6, 3), and B (6, 3).

Thus, the required area of the triangle is 18 unit2.

Question 69: A man running a racecourse notes that the sum of the distances from the two flag posts form him is always 10 m and the distance between the flag posts is 8 m. find the equation of the posts traced by the man.

ANSWER : - Let A and B be the positions of the two flag posts and P(xy) be the position of the man. Accordingly, PA + PB = 10.

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form , where a is the semi-major axis

Accordingly, 2a = 10 ⇒ a = 5

Distance between the foci (2c) = 8

⇒ c = 4

On using the relation , we obtain

Thus, the equation of the path traced by the man is .

Question 70: An equilateral triangle is inscribed in the parabola y2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

ANSWER : - Let OAB be the equilateral triangle inscribed in parabola y2 = 4ax.

Let AB intersect the x-axis at point C.

Let OC = k

From the equation of the given parabola, we have

∴The respective coordinates of points A and B are

AB = CA+ CB =

Since OAB is an equilateral triangle, OA2 = AB2.

Thus, the side of the equilateral triangle inscribed in parabola y2 = 4 ax is  .

The document Conic Sections NCERT Solutions | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
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FAQs on Conic Sections NCERT Solutions | Mathematics (Maths) Class 11 - Commerce
 1. What are conic sections?
Ans. Conic sections are the curves that are formed by the intersection of a plane with a right circular cone. The different types of conic sections include parabola, ellipse, and hyperbola.
 2. What are the applications of conic sections?
Ans. Conic sections have numerous applications in real life. For example, parabolic mirrors are used in reflecting telescopes, car headlights, and satellite dishes. Ellipses are used in the design of oval racetracks, satellite orbits, and planetary orbits. Hyperbolas are used in the design of radio antennas and in navigation systems.
 3. What is the focus of a conic section?
Ans. The focus of a conic section is a fixed point that is used to define the shape of the curve. For parabolas, the focus is a point on the axis of symmetry. For ellipses and hyperbolas, there are two foci.
 4. What is the standard equation of a parabola?
Ans. The standard equation of a parabola is y = ax^2 + bx + c, where a is a non-zero constant that determines the shape and orientation of the parabola, and (h, k) is the vertex.
 5. How are conic sections used in architecture?
Ans. Conic sections have been used in architecture for thousands of years. For example, the dome of the Pantheon in Rome is an example of a section of an ellipse. The arches in Gothic cathedrals are also based on conic sections. Additionally, conic sections are used in the design of modern architecture, such as the hyperbolic paraboloid roof of the Sydney Opera House.

## Mathematics (Maths) Class 11

156 videos|176 docs|132 tests

## Mathematics (Maths) Class 11

156 videos|176 docs|132 tests

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