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All questions of Conic Sections for JEE Exam

A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at [2008]
  • a)
    (0, 2)
  • b)
    (1, 0)
  • c)
    (0, 1)
  • d)
    (2, 0)
Correct answer is option 'B'. Can you explain this answer?

Geetika Shah answered
Vertex of a parabola is the mid point of focus and the point
where directrix meets the axis of the parabola.
Here focus is O(0, 0) and directxix meets the axis at B(2, 0)
∴ Vertex of the parabola is (1, 0)
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Let L be a normal to the parabola y2 = 4x. If L passes through the point (9, 6), then L is given by (2011)
  • a)
    y – x + 3 = 0
  • b)
    y + 3x – 33 = 0
  • c)
    y + x – 15 = 0
  • d)
    y – 2x + 12 = 0
Correct answer is option 'A,B,D'. Can you explain this answer?

Lavanya Menon answered
The equation of normal to y2 = 4x is y = mx – 2m – m3 As it passes through (9, 6)
∴ 6 = 9m – 2m – m3 ⇒ m3 – 7m + 6 = 0 ⇒ (m – 1) (m2 + m – 6) = 0
⇒ (m – 1) (m + 3) (m – 2) = 0 ⇒ m = 1, 2, –3
∴ Normal is y = x – 3 or y = 2x – 12 or y = – 3x + 33
∴ a, b, d are the correct option.

Let E1 and E2 be two ellipses whose centers are at the origin.
The major axes of E1 and E2 lie along the x-axis and the y-axis, respectively. Let S be the circle x2 + (y – 1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R respectively. Suppose that PQ = PR =
If e1 and e2 are the eccentricities of E1 and E2, respectively, then the correct expression(s) is (are)       (JEE Adv. 2015)
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A,B'. Can you explain this answer?

Krishna Iyer answered
Let E1 := 1 where a > b
and E2 := 1 where c < d
Also  S : x2 + (y – 1)2 = 2
Tangent at P(x1, y1) to S is x + y = 3
To find point of contact put x = 3 – y in S.
We get P(1, 2) Writing eqn of tangent in parametric form
 and 
 and 
eqn of tangent to E1 at Q is
which is identical to
⇒ a2 = 5 and b2 = 4 ⇒ 
eqn of tangent to E2 at R is
 identical to 
⇒ c2 = 1, d2 = 8 ⇒  
 

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)
  • a)
    1/8
  • b)
    8
  • c)
    4
  • d)
    1/4
Correct answer is option 'C'. Can you explain this answer?

Geetika Shah answered
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

The locus of the vertices of the family of parabol as     [2006]
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Rohit Jain answered
Given parabola is  
∴ Vertex of parabola is 
To find locus of this vertex,
 64xy = 105
which is the required locus.

An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)
  • a)
    equation of ellipse is x2 + 2y2 = 2
  • b)
    the foci of ellipse are (±1, 0)
  • c)
    equation of ellipse is x2 + 2y2 = 4
  • d)
    the foci of ellipse are (±, 0)
Correct answer is option 'A,B'. Can you explain this answer?

Pioneer Academy answered
The given hyperbola is
...(1)
which is a rectangular hyperbola (i.e.  a = b)
Let the ellipse be 
Its eccentricity  = 
So, the equation of ellipse becomes x2 + 2 y2 = a2 ...(2)
Let  the hyperbola (1) and ellipse (2) intersect each other at P( x1 ,y1).
Then slope of hyperbola (1) at P is given by
and that of ellipse (2) at P is
As the two curves intersect orthogonally,
∴ m1m2 =-1
Also P ( x1,y1) lies on x2 -y2 =
...(ii)
Solving (i) and (ii), we get 
Also P ( x1,y1) lies on ellipse x2 + 2y2 = a2
 1 + 1 = a2 or a2 =2
∴ The required ellipse is x2 + 2y2 = 2 whose foci

The ellipse  x2 + 4y2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is : [2009]
  • a)
    x2 + 12y2= 16
  • b)
    4 x2 + 48y2= 48
  • c)
    4 x2 + 64y2 = 48
  • d)
    x2 + 16y2= 16
Correct answer is option 'A'. Can you explain this answer?

Anushka Ahuja answered
Solution:

Given that the ellipse x^2/4 + y^2 = 1 is inscribed in a rectangle aligned with the coordinate axes and this rectangle is inscribed in another ellipse that passes through the point (4, 0).

Let's find the equation of the outer ellipse.

Step 1: Find the coordinates of the four corners of the rectangle.
The corners of the rectangle can be obtained by solving the equation of the ellipse with x = ±2 and y = ±1.
For x = 2, we have 4/4 + y^2 = 1, which gives y = ±√3.
So, the coordinates are (2, √3) and (2, -√3).
Similarly, for x = -2, we have 4/4 + y^2 = 1, which gives y = ±√3.
So, the coordinates are (-2, √3) and (-2, -√3).

Step 2: Find the equation of the ellipse passing through the points (2, √3), (2, -√3), (-2, √3), and (-2, -√3).
Using the standard form of an ellipse, the equation is:
(x - h)^2/a^2 + (y - k)^2/b^2 = 1, where (h, k) is the center of the ellipse, and a and b are the semi-major and semi-minor axes, respectively.

Step 3: Find the center of the ellipse.
The center of the ellipse is the midpoint of the diagonals of the rectangle.
The midpoint of the diagonals is ((2 + (-2))/2, (√3 + (-√3))/2) = (0, 0).

Step 4: Find the semi-major and semi-minor axes.
The semi-major axis is the distance from the center to one of the corners of the rectangle, which is 2.
The semi-minor axis is the distance from the center to one of the sides of the rectangle, which is 1.

Therefore, the equation of the outer ellipse is:
(x - 0)^2/2^2 + (y - 0)^2/1^2 = 1
Simplifying, we get x^2/4 + y^2 = 1.

Hence, the correct answer is option A) x^2 - 12y^2 = 16.

 The angle between the tangents drawn from the point (1, 4) to the parabola y2 = 4x is (2004S)
  • a)
    π/6
  • b)
    π/4
  • c)
    π/3
  • d)
    π/2
Correct answer is option 'C'. Can you explain this answer?

Jay Datta answered
To find the angle between the tangents drawn from a point to a curve, we can first find the points of tangency and then use the formula for the angle between two lines.

The given parabola is y^2 = 4x. Let's find the equation of the tangent lines.

Using implicit differentiation, we can find the derivative of the equation of the parabola:

2y dy/dx = 4

dy/dx = 2/y

At any point (x, y) on the parabola, the slope of the tangent line is given by dy/dx = 2/y.

Now let's find the points of tangency by substituting the coordinates of the given point (1, 4) into the equation of the curve:

4^2 = 4(1)
16 = 4

This means that the point (1, 4) lies on the parabola, so it is one of the points of tangency.

Next, let's find the equation of the tangent line at (1, 4). Since the slope of the tangent line is dy/dx = 2/y, we can substitute y = 4 into this equation:

dy/dx = 2/4 = 1/2

So the slope of the tangent line at (1, 4) is 1/2.

Using the point-slope form of a line, we can find the equation of the tangent line:

y - 4 = (1/2)(x - 1)

Simplifying, we get:

2y - 8 = x - 1
2y = x + 7

Now let's find the other point of tangency. We can solve the equation of the parabola for x in terms of y and then substitute the equation of the tangent line into the parabola equation:

y^2 = 4x
x = y^2/4

Substituting the equation of the tangent line into the parabola equation, we get:

2y = (y^2/4) + 7

Multiplying through by 4 to eliminate the fraction, we get:

8y = y^2 + 28

Rearranging, we get a quadratic equation:

y^2 - 8y + 28 = 0

Using the quadratic formula, we can find the solutions for y:

y = (8 ± sqrt((-8)^2 - 4(1)(28))) / 2

y = (8 ± sqrt(64 - 112)) / 2
y = (8 ± sqrt(-48)) / 2
y = (8 ± 4i√3) / 2
y = 4 ± 2i√3

Since we are given that the point (1, 4) is one of the points of tangency, the other point of tangency must be (1, 4 - 2i√3).

Now let's find the slope of the tangent line at this point. Using the same process as before, we find that the slope of the tangent line is -1/2.

Therefore, the equation of the other tangent line is:

y - (4 - 2i√3) = (-1/2)(x - 1)

Simplifying, we get:

2y - (8

If x + y = k is normal to y2 = 12 x, then k is (2000S)
  • a)
    3
  • b)
    9
  • c)
    –9
  • d)
    –3
Correct answer is option 'B'. Can you explain this answer?

Gaurav Rane answered
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

Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are [2002]
  • a)
    x = ±( y+ 2a)
  • b)
    y = ± ( x+ 2a)
  • c)
    x = ± ( y+a)
  • d)
    y = ± ( x+a)
Correct answer is option 'B'. Can you explain this answer?

Juhi Nambiar answered
Any tangent to the parabola y2 = 8ax is...(i)
If (i) is a tangent to the circle, x2 + y2 = 2a2 then,
 
⇒ m2(1 + m2) = 2 ⇒ (m2 + 2)(m2 – 1) = 0 ⇒ m = ± 1.
So from (i), y = ± (x + 2a).

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)
  • a)
    a hyperbola
  • b)
    a parabola
  • c)
    an ellipse
  • d)
    a straight line
Correct answer is option 'D'. Can you explain this answer?

Saranya Choudhary answered
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.

The equation of the common tangent to the curves y2 = 8x and xy = –1 is  (2002S)
  • a)
    3y = 9x + 2
  • b)
    y = 2x + 1
  • c)
    2y = x + 8
  • d)
    y = x + 2
Correct answer is option 'D'. Can you explain this answer?

Nisha Rane answered
To find the equation of the common tangent to the curves, we need to find the point of intersection of the two curves and then find the slope of the tangent at that point.

First, let's find the point of intersection by solving the two equations simultaneously:

y^2 = 8x ...(1)
xy = 4 ...(2)

From equation (2), we can express x in terms of y:
x = 4/y

Substituting this value of x in equation (1), we get:
y^2 = 8(4/y)
y^2 = 32/y
y^3 = 32

Taking the cube root of both sides, we get:
y = 2

Substituting this value of y back into equation (2), we get:
x = 4/2
x = 2

So the point of intersection is (2, 2).

To find the slope of the tangent at this point, we can differentiate both equations with respect to x:

Differentiating equation (1) implicitly, we get:
2y(dy/dx) = 8
dy/dx = 8/(2y)
dy/dx = 4/y

Differentiating equation (2) implicitly, we get:
y + x(dy/dx) = 0
dy/dx = -y/x

Substituting the coordinates of the point of intersection (2, 2), we get:
dy/dx = 4/2 = 2
dy/dx = -2/2 = -1

The slopes of the tangents to the curves at the point of intersection are 2 and -1, respectively.

Now, let's use the point-slope form of a line to find the equation of the tangent with slope 2:

Using the point (2, 2) and slope 2, the equation of this tangent is:
y - 2 = 2(x - 2)

Simplifying, we get:
y - 2 = 2x - 4
y = 2x - 2

Similarly, using the point-slope form of a line to find the equation of the tangent with slope -1:

Using the point (2, 2) and slope -1, the equation of this tangent is:
y - 2 = -1(x - 2)

Simplifying, we get:
y - 2 = -x + 2
y = -x + 4

So the equations of the common tangents to the curves y^2 = 8x and xy = 4 are y = 2x - 2 and y = -x + 4.

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)
  • a)
    3
  • b)
    6
  • c)
    9
  • d)
    15
Correct answer is option 'D'. Can you explain this answer?

Tushar Saha answered
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

The equation 2x2 + 3y2 – 8x – 18y + 35 = k represents (1994)
  • a)
    no locus if k > 0
  • b)
    an ellipse if k < 0
  • c)
    a point if k = 0
  • d)
    a hyperbola if k > 0
Correct answer is option 'C'. Can you explain this answer?

Deepika Sen answered
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).

If P = (x, y), F1 = (3, 0), F2 = (–3, 0) and 16x2 + 25y2 = 400, then PF1 +PF2 equals          (1998 - 2 Marks)
  • a)
    8
  • b)
    6
  • c)
    10
  • d)
    12
Correct answer is option 'C'. Can you explain this answer?

Chirag Verma answered
The ellipse can be written as,
Here a2 = 25, b2 = 16, but b2 = a2 (1 – e2)
⇒ 16/25 = 1– e2 ⇒ e2 =1 – 16/25 = 9/25  ⇒ e = 3/5
Foci of the ellipse are (+ ae, 0) = (± 3, 0), i.e., F1 and F2
∴ We have PF1 + PF2 = 2a = 10 for every point P on the ellipse.

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)
  • a)
    9x2 – 8y2 + 18x – 9 = 0
  • b)
    9x2 – 8y2 – 18x + 9 = 0
  • c)
    9x2 – 8y2 – 18x – 9 = 0
  • d)
    9x2 – 8y2 + 18x + 9 = 0  
Correct answer is option 'B'. Can you explain this answer?

Anaya Patel answered
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

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)
  • a)
    {–1, 1}
  • b)
    {–2, 2}
  • c)
    {–2, –1/2}
  • d)
    {2, –1/2}
Correct answer is option 'A'. Can you explain this answer?

Shruti Khanna answered
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

If the line = 2 touches the hyperbola x2 – 2y2 = 4, then the point of contact is (2004S)
  • a)
    (– 2, √6)
  • b)
    (– 5, 2√6)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Anand Kumar answered
Write the general equation of the tangent(at to the hyperbola and compare the coefficient of this equation with the given equation as A1/A2=B1/B2=C1/C2.(where A1,A2,B1,B2,C1,C2 are coefficient of both lines)

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)
  • a)
    (–6, –11)
  • b)
    (–9, –13)
  • c)
    (–10, –15)
  • d)
    (–6, –7)
Correct answer is option 'D'. Can you explain this answer?

Shraddha Patel answered
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).

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)
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Niti Gupta answered
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 

The circle C1 : x2 + y2 = 3, with centre at O, intersects the parabola x2 = 2y at the point P in the first quadrant. Let the tangent to the circle C1, at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 2 and centres Q2 and Q3, respectively. If Q2 and Q3 lie on the y–axis, then (JEE Adv. 2016)
  • a)
    Q2Q3 = 12
  • b)
    R2R3 =
  • c)
    area of the triangle OR2R3 is
  • d)
    area of the triangle PQ2Q3 is
Correct answer is option 'A,B,C'. Can you explain this answer?

Mansi Menon answered
C1 : x2 + y2 = 3 ..(i)
parabola : x2  = 2y ...(ii)
Intersection point of (i) and (ii) in first quadrant y2 + 2y – 3 = 0 ⇒ y = 1 (∵y ≠ -3)
∴x =
P( ,1)
Equation of tangent to circle C1 at P is x + y-3 = 0
Let centre of circle C2 be (0, k); r = 2
 k = 9 or –3
∴ Q2 (0, 9), Q3 (0, –3)
(a)Q2 Q3 = 12
(b) R2R3 = length of transverse common tangent
(c) Area  × length of ⊥ from originto tangent
(d) Area  distance of P from
y–axis =

PASSAGE 3
Tangents are drawn from the point P(3, 4) to the ellipse  touching the ellipse at points A and B. (2010)
Q. The orthocenter of the triangle PAB is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Aryan Sharma answered
Let H be the  orthocentre of ΔPAB, then as BH ⊥ AP, BH is a horizontal line  through B.
∴ y- cordinate of B = 8/5
Let H has coordinater (α, 8 5)
Then slope of PH
and slope of AB 
But PH ⊥ AB ⇒ 
⇒ 4 = –5α + 15 or α = 11/5
Hence

PASSAGE 6
Let F1(x1, 0) and F2(x2, 0) for x1 < 0 and x2 > 0, be the foci of the ellipse   Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.
Q. If the tangen ts to the ellipse at M an d N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is   (JEE Adv. 2016)
  • a)
    3 : 4
  • b)
    4 : 5
  • c)
    5 : 8
  • d)
    2 : 3
Correct answer is option 'C'. Can you explain this answer?

Pranavi Ahuja answered
For ellipse
∴ F1(– 1, 0) and F2 (1, 0) Parabola with vertex at (0, 0) and focus at F2(1, 0) is y2 = 4x.
Intersection points of ellipse and parabola are and 
Tangents to ellipse at M and N are
and 
Their intersection point is R (6, 0)
Also normal to parabola at  is 
Its intersection with x-axis is 
Now ar (ΔMQR)  =
Also area (MF1NF2) = 2 × Ar (F1MF2)

The  normal at the point   (bt12 , 2bt1)  on a parabola meets the parabola again in the point (bt22, 2bt2) , then            [2003]
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Tanishq Choudhary answered
Equation of the normal to a parabola y2 = 4bx at point
(bt12,2bt1) is y = – t1 x + 2bt1+ bt13
As given, it also passes through (bt22 ,2bt2) then
 
 
 ⇒ t2 + t1 =

Angle between the tangents to the curve y = x2 - 5x+6 at the points (2, 0) and (3, 0) is [2006]
  • a)
    π
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Akshita Roy answered
     ∴ m1 = (2x - 5)(2, 0) =-1 ,
m2 = (2x - 5)(3, 0)=1 ⇒ m1m2 =-1
i.e. the tangents are perpendicular to each other.

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