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# Test: Single Correct MCQs: Conic Sections | JEE Advanced

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## 35 Questions MCQ Test Mathematics For JEE | Test: Single Correct MCQs: Conic Sections | JEE Advanced

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Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 1

### The equation    represents(1981 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 1

Given that

As r > 1

∴ 1 – r < 0  and 1 + r > 0

∴ Let  1 – r = – a2, 1 + r = b2, then we get

which is not possible for any real values of x and y.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 2

### Each of the four inequalties given below defines  a region in the xy plane. One of these four regions does not have the following property. For any two points (x1, y1) and (x2, y2) in th e region , the point    is also in the region. The inequality defining this region is (1981 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 2

(a) x2 + 2y2 ≤ 1 represents interior region of an ellipse where on taking any two pts the mid pt of that segment will also lie inside that ellipse
(b) Max { | x |, | y | } ≤  1 ⇒ | x | ≤ 1, | y | ≤ 1 ⇒ – 1 ≤ x ≤ 1 and – 1 ≤ y ≤ 1
which represents the interior region of a square with its sides x = ± 1 and y = ± 1 in which for any two pts, their mid pt also lies inside the region.
(c) x2 – y2 ≥ 1 repr esents the exterior r egion of hyperbola in which if we take two points (2, 0) and (– 2, 0) then their mid pt (0, 0) does not lie in the same region (as shown in the figure.)

(d) y2 ≤ x represents interior region of parabola in which for any two pts, their mid point also lie inside the region.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 3

### The equation 2x2 + 3y2 – 8x – 18y + 35 = k represents (1994)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 3

We have 2x2 + 3y2 – 8x – 18y + 35 = k
⇒ 2 (x – 2)2 + 3 (y – 3)2 = k
For k = 0, we get 2 (x – 2)2 + 3 (y – 3)2 = 0 which represents the point (2, 3).

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 4

Let E be the ellipse  and C be the circle x2 + y2 = 9. Let P and Q be the points (1, 2) and (2, 1) respectively. Then (1994)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 4

Since 12 + 22 = 5 < 9 and 22 + 11 = 5 < 9 both P and Q lie inside C.

Also  and   P lies outside E and Q lies inside E. Thus P lies inside C but outside E.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 5

Consider a circle with its centre lying on the focus of the parabola y2 = 2px such that it touches the directrix of the parabola. Then a point of intersection of the circle and parabola is (1995S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 5

The focus of parabola y2 = 2px is   and directrix
x = – p/2

In the figure, we have supposed that p > 0]

∴ Centre of circle is  and radius

∴ Equation of circle is

For pts of intersection of y2 = 2px ...(i)
and 4x2 + 4y2 – 4px – 3p2 = 0 ...(ii)
can be obtained by solving (i) and (ii) as follows 4x2 + 8px – 4px – 3p2 = 0 ⇒ (2x + 3p) (2x – p) = 0

⇒ y2 = – 3p2 (not possible), p2 ⇒ y = + p

∴ Required pts are (p/2, p), (p/2, – p)

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 6

The radius of the circle passing through the foci of theellipse  and having its centre at (0, 3) is (1995S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 6

For ellipse   = 1, a = 4, b = 3

∴ Foci are

Centre of circle is at (0, 3) and it passes through

, therefore radius of circle =

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 7

Let P (a secθ, b tanθ) and Q (a secφ, b tanφ), where q + f = π/ 2, be two points on the hyperbolaIf (h, k) is the point of intersection of the normals at P and Q, then k is equal to (1999 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 7

KEY CONCEPT : Equation of the normal to the hyperbola

at the point (a secα, b tanα) is given by ax cosα + by cotα = a2 + b2

Normals at

where   and these pass through (h, k)
∴ ah cosθ + bk cotθ = a2 + b2 ah sinθ + bk tanθ = a2 + b2
Eliminating h, bk (cotθ sinθ – tan cosθ ) = (a2 + b2)
(sinθ – cosθ ) or k = – (a2 + b2)/b

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 8

If x = 9 is the chord of contact of the hyperbola x2 – y2 = 9, then the equation of the corresponding pair of tangents is (1999 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 8

Chord x = 9 meets x2 – y2 = 9 at (9,6 ) and (9, –6)
at which tangents are 9x –6y =9 and 9x+6y=9
or 3x –2y – 3 =0 and 3x + 2y – 3=0
∴ Combined equation of tangents is (3x –2y –3)(3x + 2y – 3)=0
or9x2 – 8y2 – 18x + 9 = 0

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 9

Th e curve described par ametr ically by x = t2 + t + 1, y = t2 – t + 1 represents (1999 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 9

KEY CONCEPT
The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a parabola if Δ ≠ 0 and h2 = ab where Δ = abc + 2fgh – af 2 – bg2 – ch2
Now we have x = t2 + t + 1 and  y = t2 – t + 1

(Adding and subtracting values of x and y)
Eliminating t,   2 (x + y) = (x – y)2 + 4 ......... (1)
⇒ x2 – 2xy + y2 – 2x – 2y + 4  = 0 ......... (2)
Here, a = 1, h = – 1, b = 1, g = – 1, f = – 1, c = 4
∴Δ ≠ 0. and h2 = ab
Hence the given curve represents a parabola.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 10

If x + y = k is normal to y2 = 12 x, then k is (2000S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 10

y = mx + c is normal to the parabola y2 = 4 ax if c = – 2am – am3
Here m = –1, c = k and a = 3
∴  c = k = – 2 (3) (–1) – 3 (–1)3 = 9

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 11

If the line x – 1 = 0 is the directrix of the parabola y2 – kx + 8 = 0, then one of the values of k is (2000S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 11

KEY CONCEPT : The directrix of the parabola y2 = 4a (x – x1) is given by x = x1 – a.

y2 = kx – 8 ⇒ y2 =

Directrix of parabola is  x =

Now, x = 1 also coincides with  =

On comparision,  =1,  or k2 - 4k - 32 = 0
On solving  we get k = 4

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 12

The equation of the common tangent touching the circle (x -3)2 + y2 = 9 and the parabola y2 = 4x above the x-axis is (2001S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 12

Let the equation of tangent to y2 = 4x be y =

where m is the slope of the tangent.
If it is tangent to the circle (x – 3)2 + y2 = 9 then length of perpendicular to tangent from centre (3, 0) should be equal to the radius 3.

∴ Tangents are x –y + 3=0 and

x +y+ 3 = 0 out of which x –y + 3=0 meets the parabola at (3, 2) i.e., above x-axis.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 13

The equation of the directrix of the parabola y2 + 4y + 4x + 2 = 0 is  (2001S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 13

y2 + 4y + 4x + 2 = 0
y2 + 4y + 4 = – 4x + 2 (y + 2)2 = –4 (x – 1/2)
It is of the form Y2 = –  4AX
whose directrix is given by X = A
∴ Req. equation is x – 1/2 = 1 ⇒ x = 3/2.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 14

If a > 2b > 0 then the positive value of m for which is a common tangent to x2 + y2 = b2  and (x – a)2 + y2 = b2 is          (2002S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 14

Given that a > 2b > 0 and m > 0
Also ...(1)
is tangent to x2 + y2 = b2 ...(2)
as well as to (x – a)2 + y2 = b2 ...(3)
∵ (1) is tangent to (3)

[length of perpendicular from (a, 0) to (1) = radius b]

or am = 0 (not possible as a, m > 0)

(∵ m > 0)

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 15

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola y2 = 4ax is another parabola with directrix  (2002S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 15

If (h, k) is the mid point of line joining focus (a, 0) and Q (at2, 2at) on parabola then k = at

Eliminating t, we get  2h =

⇒ k2 = a (2h – a) ⇒  k2 = 2a (h – a/2)

∴ Locus of (h, k) is y2 = 2a (x – a/2)

whose directrix is (x – a/2) =

⇒ x = 0

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 16

The equation of the common tangent to the curves y2 = 8x and xy = –1 is  (2002S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 16

The given curves are y2 = 8x ...(1)
and xy = – 1 ...(2)
If m is the slope of tangent to (1), then eqn of tangent is y = mx + 2/m
If this tangent is also a tangent to (2), then putting value of y in curve (2)

= 0 ⇒ m2 x2 + 2x + m = 0

We should get repeated roots for the eqn (condition of tangency)
⇒ D = 0
∴ (2)2 – 4m2.m = 0 ⇒ m3 = 1 ⇒ m = 1
Hence required tangent is y = x + 2

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 17

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse  is

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 17

The given ellipse is

Then a2 = 9, b2 = 5  ⇒ e

∴ end point of latus rectum in first quadrant is L (2, 5/3)

Equation of tangent at L is

It meets x-axis at A (9/2, 0) and y-axis at B (0, 3)

∴ Area of ΔOAB =

By symmetry area of quadrilateral = 4 × (Area ΔOAB) =  = 27 sq. units.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 18

The focal chord to y2 = 16x is tangent to (x – 6)2 + y2 = 2, then the possible values of the slope of this chord, are (2003S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 18

For parabola y2 = 16x, focus ≡ (4, 0).
Let m be the slope of focal chord then eqn is y = m (x – 4) ...(1)
But given that above is a tangent to the circle (x – 6)2 + y2 = 2
With Centre, C (6, 0), r = 2
∴ Length of ⊥ lar from (6, 0) to (1) = r

⇒ 2m2 = m2 + 1 ⇒ m2 = 1  ⇒ m = ±1

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 19

For hyperbola  which of the following remains constant with change in ‘α’ (2003S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 19

The given eqn of hyperbola is

⇒ a = cosα, b = sinα

⇒ ae = 1
∴ foci ( + 1 , 0)
∴ foci remain constant with respect to α.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 20

If tangents are drawn to the ellipse x2 + 2y2 = 2, then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is (2004S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 20

Any tangent to ellipse

⇒ 2h = 2 secθ and 2k = cosecθ (Using mid pt. formula)

Required locus,

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 21

The angle between the tangents drawn from the point (1, 4) to the parabola y2 = 4x is (2004S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 21

y = mx + 1/m
Above tangent passes through (1, 4) ⇒ 4 = m + 1/m ⇒  m2 – 4m + 1 = 0
Now angle between the lines is given by

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 22

If the line = 2 touches the hyperbola x2 – 2y2 = 4, then the point of contact is (2004S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 22

Equation of tangent to hyperbola x2 – 2y2 = 4 at any point (x1, y1) is xx1 – 2yy1 = 4
Comparing with 2x +y = 2 or 4 x + 2y=4
⇒ x1 = 4 and  – 2y1 = 2 ⇒ (4, – ) is the required point.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 23

The minimum area of triangle formed by the tangent to the  & coordinate axes is (2005S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 23

Any tangent to the ellipse

at P (a cosθ, b sinθ) is

It meets co-ordinate axes at A (a secθ, 0) and B (0, b cosecθ)
∴ Area of ΔOAB = x a secθ × b cosecθ

For Δ to be min, sin 2θ should be max. and we know max value of sin 2θ= 1

∴ Δ max = ab sq. units.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 24

Tangent to the curve y = x2 + 6 at a point (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are (2005S)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 24

The given curve is y = x2 + 6 Equation of tangent at (1, 7) is (y + 7) = x .1 + 6
⇒ 2x – y + 5 = 0 ...(1)
As given this tangent (1) touches the circle x2 + y2 +16x + 12y + c = 0 at Q
Centre of circle = (– 8, – 6).

Then equation of CQ which is perpendicular to (1) and passes through (– 8, – 6) is y + 6 = -(x + 8)

⇒ x + 2y + 20 = 0 ...(2)
Now Q is pt. of intersection of (1) and (2)
∴   Solving eqn (1) & (2) we get x = – 6, y = – 7
∴ Req. pt. is (– 6, – 7).

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 25

The axis of a parabola is along the line y = x and the distances of its vertex and focus from origin are and respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is (2006 - 3M, –1)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 25

Since, distance of vertex from origin is and focus is
∴ Vertex is (1, 1) and focus is (2, 2), directrix  x + y = 0

∴ Equation of parabola is

( x – 2)2 + (y – 2)2

⇒ 2(x2 - 4x+ 4)+ 2(y2 - 4y + 4) = x2 + y2+ 2xy
⇒ x2 + y2 – 2xy = 8 (x + y – 2)
⇒ (x – y)2 = 8 (x + y – 2)

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 26

A hyperbola, having the transverse axis of length 2 sinθ, is confocal with the ellipse 3x2 + 4y2 = 12. Then its equation is (2007 - 3 marks)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 26

The length of transverse axis = 2 sinθ = 2a
⇒ a = sinθ
Also for ellipse 3x2 + 4y2 = 12

∴ Focus of ellipse =

As hyperbola is confocal with ellipse, focus of hyperbola = (1, 0)  ⇒ ae = 1 ⇒ sinθ × e = 1 ⇒ e = cosecθ

∴ b2 = a2 (e2 – 1) = sin2θ (cosec2θ – 1) = cos2θ

∴ Equation of hyperbola is

or, x2cosec2θ – y2 sec2θ = 1

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 27

Let a and b be non-zero real numbers. Then, the equation (ax2 + by2 + c) (x2 – 5xy + 6y2) = 0 represents (2008)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 27

x2 – 5xy + 6y2 = 0 represents a pair of straight lines given by x – 3y = 0 and x – 2y = 0.
Also ax2 + by2 + c = 0 will represent a circle if a = b and c is of sign opposite to that of a.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 28

Consider a branch of the hyperbola with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is (2008)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 28

The given hyperbola is

∴ a = 2, b =

Clearly ΔABC is a right triangle.

=

=

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 29

The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x2 + 9y2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and the origin O is (2009)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 29

The given ellipse is x2 + 9y= 9 or

So, that A( 3,0) and B (0,1)

∴ Equation of AB is

or x + 3y – 3 = 0 (1)
Also auxillary circle of given ellipse is x2 +y2= 9 (2)
Solving equation (1) and (2), we get the point M where line AB meets the auxillary circle.
Putting  x = 3 – 3y from eqn (1) in eqn (2) we get (3 - 3 y)2 +y= 9
⇒ 9 -18y + 9y2 +y= 9 ⇒ 10y2 -18y=0

Clearly M

∴ Area of ΔOAM =

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 30

The normal at a point P on the ellipse x2 + 4y2 = 16 meets the x - axis at Q. If  M is the mid point of the line segment PQ, then the locus of  M  intersects the latus rectums of the given ellipse at the points (2009)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 30

The given ellipse is

such that a2 = 16 and b2 = 4

Let P(4cosθ, 2sinθ) be any point on the ellipse, then equation of normal at P is

4 x sinθ - 2y cosθ = 12 sinθ cosθ

∴ Q, the point where normal at P meets x –axis, has coordin ates (3 cosθ, 0)

∴ Mid point of PQ is M

For locus of point M we consider

and y = sinθ

and sinθ= y

...(1)
Also the latus rectum of given ellipse is

Solving equations (1) and (2), we get

∴ The required points are

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 31

The locus of the orthocentre of the triangle formed by the lines
(1+ p) x – py + p (1+ p) = 0,
(1+ q) x – qy + q (1+ q) = 0,
and y = 0, where p ≠ q, is (2009)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 31

The triangle is formed by the lines

AB : (1 + p) x - py + p (1 +p)= 0
AC : (1+ q)x -qy +q(1+ q)= 0
BC :y =0
So that the vertices are A( pq, ( p + 1)(q+ 1)), B(- p, 0), C (-q, 0)
Let H (h,k) be the orthocentre of Δ ABC.
Then as AH ^ BC and passes through A( pq, ( p + 1)(q+ 1)) The eqn of AH is x = pq
∴ h = pq ...(1)
Also BH is perpendicular to AC

(using eqn  (1)
⇒ k =- pq ...(2)
From (1) and (2) we observe  h + k = 0
∴ Locus of (h, k) is x + y = 0 which is a straight line.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 32

Let P(6, 3) be a point on the hyperbola  If the
normal at the point P intersects the x-axis at (9, 0), then the eccentricity of the hyperbola is (2011)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 32

For hyperbola

, we have

∴ Slope of normal at P (6, 3)

∴ Equation of normal is

As it intersects x-axis at (9, 0)

...(1)
Also for hyperbola, b2 = a2 (e2 – 1)
Using a2 = 2b2; we get b2 = 2b2 (e2 – 1)

or

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 33

Let (x, y) be any point on the parabola y2 = 4x. Let P be the point that divides the line segment from (0, 0) to (x, y) in the ratio 1 : 3. Then the locus of P is (2011)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 33

Let A (x, y) = (t2, 2 t) be any point on parabola y2 = 4x.

Let P (h, k) divides OA in the ratio 1 : 3

Then (h, k) =

∴ locus of P (h, k) is x = y2.

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 34

The ellipse   is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E2  is (2012)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 34

As rectangle ABCD circumscribed the ellipse

∴ A = (3, 2)

Let the ellipse circumscribing the rectangle ABCD is

Given that it passes through (a, 4)
∴ b2 = 16
Also it passes through A (3, 2)

Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 35

The common tangents to the circle x2 + y2 = 2 and the parabola y2 = 8x touch the circle at the points P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is(JEE Adv. 2014)

Detailed Solution for Test: Single Correct MCQs: Conic Sections | JEE Advanced - Question 35

Let the tangent to y2 = 8x be  y =

If it is common tangent to parabola and circle, thenis a tangent to x2 + y2 = 2

⇒ m4 + m2 – 2 = 0 ⇒ (m2 + 2) (m2 – 1) = 0 ⇒ m = 1 or – 1
∴ Required tangents are y = x + 2 and y = – x – 2

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## Mathematics For JEE

130 videos|359 docs|306 tests