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NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

Question 1: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

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

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

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 2: 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

NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

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 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections.

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

Therefore, the arch is in the form of a parabola whose equation is NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections.

When y = 2 m, NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

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

Question 3: 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 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections.

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,

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

∴Equation of the parabola, NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections or 6x2 = 625y

NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

The x-coordinate of point D is 18.

Hence, at x = 18,

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

∴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 4: 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 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections, where a is the semi-major axis

Accordingly, 2a = 8 ⇒ a = 4

b = 2

Therefore, the equation of the semi-ellipse is NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

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.

NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

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

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

∴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 5: 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.

NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

In ΔPBQ, NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

In ΔPRA, NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

Thus, the equation of the locus of point P on the rod is NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections.

 

Question 6: 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).

NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

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

 

Question 7: 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(x, y) 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.

NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

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 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections, where a is the semi-major axis

Accordingly, 2a = 10 ⇒ a = 5

Distance between the foci (2c) = 8

c = 4

On using the relation NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections, we obtain

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

Thus, the equation of the path traced by the man is NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections.

 

Question 8: 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 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

∴The respective coordinates of points A and B are NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

AB = CA+ CB = NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

Since OAB is an equilateral triangle, OA2 = AB2.

NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

 NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

Thus, the side of the equilateral triangle inscribed in parabola y2 = 4 ax is  NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections.

The document NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on NCERT Solutions Class 11 Maths Chapter 10 - Conic Sections

1. What are conic sections?
Ans. Conic sections are the curves formed when a cone is sliced by a plane. They include the shapes of a circle, ellipse, parabola, and hyperbola.
2. How are conic sections used in real life?
Ans. Conic sections have various applications in real life. For example, the shapes of satellite dishes, mirrors, and lenses are based on conic sections. They are also used in designing roller coasters, telescopes, and even in analyzing the orbits of planets.
3. What is the equation of a circle in conic sections?
Ans. The equation of a circle in the coordinate plane is given by (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
4. How do conic sections relate to algebraic equations?
Ans. Conic sections can be represented by algebraic equations. Each type of conic section has a specific equation that relates the coordinates (x, y) on the curve. By solving these equations, we can determine the properties and characteristics of the conic sections.
5. How can one identify the type of conic section from its equation?
Ans. The type of conic section can be determined by analyzing the coefficients and constants in its equation. For example, if the equation is of the form Ax^2 + By^2 = C, then it represents an ellipse if A and B have the same sign, a hyperbola if they have opposite signs, and a circle if A = B. Similarly, equations of the form y = ax^2 + bx + c represent parabolas.
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