Integer Answer Type Questions: Conic Sections | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

1. The line 2x + y = 1 is tangent to the hyperbola 
 If this line passes through the point of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is (2010) 

Ans.  (2)

Sol.  Intersection point of nearest directrix x = and x-axis

is 

As 2x +y=1 passes through

Also y = –2x+1 is a tangent to 

⇒ 4a²– a²e² – 1=1 ⇒  

⇒ 4e² –e⁴ +e² = 4 ⇒ e⁴ – 5e⁴ + 4=0

⇒ e² = 4 ase>1 for hyperbola.  ⇒ e= 2

2. Consider the parabola y² = 8x . Let Δ₁ be the area of the triangle formed by the end points of its latus rectum and the pointon the parabola and Δ₂ be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum. Then  is     (2011) 

Ans. (2)

Sol. Δ₁ =  Area of ΔPLL'  

Equation of AB,  y = 2x + 1 Equation of AC, y = x + 2 Equation of BC,  – y = x + 2 Solving above equations we get A (1, 3), B (–1, –1), C (–2, 0)

3. Let S be the focus of the parabola y²  = 8x and let PQ be the common chord of the circle x² + y² – 2x – 4y =  0 and the given parabola. The area of the triangle PQS is     (2012)

Ans.  (4)

Sol.  We observe both parabola y² = 8x and circle x² + y² – 2x – 4y = 0 pass through origin
∴ One end of common chord PQ is origin. Say P(0, 0)
Let Q be the point (2t², 4t), then it will satisfy the equation of circle.
∴ 4t⁴ + 16t² – 4t² – 16t = 0 ⇒ t4 + 3t² – 4t = 0  ⇒ t (t³ + 3t – 4) = 0
⇒ t (t – 1)(t² + t – 4) = 0   ⇒ t = 0 or 1
For t = 0,  we get point P, therefore t = 1 gives point  Q as (2, 4).
We also observe here that P(0, 0) and Q(2, 4) are end points of diameter of the given circle and focus of the parabola is the point S(2, 0).

∴ Area of  ΔPQS sq. units

4. A vertical line passing through the point (h, 0) intersects the
 ellipse 

at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If Δ(h) = area of the triangle   then

(JEE Adv. 2013)
 (a) g(x) is continuous but not differentiable at a
 (b) g(x) is differentiable on R
 (c) g(x) is continuous but not differentiable at b
 (d) g(x) is continuous and differentiable at either (a) or (b) but not both

Ans. (9)

Sol. Vertical line x = h, meets the ellipse  at

and 

By symmetry, tangents at P and Q will meet each other at x-axis.

Tangent at P is  

which meets x-axis at

Area of  ΔPQR  = 

i.e.,

∴ Δ(h) is a decreasing function.

= 45 – 36 = 9

5. If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are tangents to the circle (x – 3)² + (y + 2)² = r², then the value of r² is (JEE Adv. 2015)

Ans. (2)

Sol. End points of latus rectum of y² = 4x are (1, +2)
Equation of normal to y² = 4x at (1, 2) is y – 2 = –1(x – 1)
or x + y –3 = 0
As it is tangent to circle (x – 3)² + (y + 2)² = r²

6. Let the curve C be the mirror image of the parabola y² = 4x with respect to the line x + y + 4 = 0. If A and B are the points of intersection of C with the line  y = –5, then the distance between A and B is (JEE Adv. 2015)

Ans. (4) 

Sol. Let (t², 2t) be any point on y² = 4x. Let (h, k) be the image of (t², 2t) in the line x + y + 4 = 0. Then

⇒ h = –(2t + 4) and k = –(t² + 4)
For its intersection with, y = –5, we have –(t² + 4) = –5 ⇒ t = +1
∴ A(–6, –5) and B(–2, –5) ∴ AB = 4.

7. Suppose that the foci of the ellipse

 = 1 are (f₁, 0)and (f₂, 0) where f₁ > 0 and f₂ < 0. Let P₁ and P₂ be two parabolas with a common vertex at (0, 0) and with foci at (f₁, 0) and (2f₂, 0), respectively. Let T₁ be a tangent to P₁ which passes through (2f₂, 0) and T₂ be a tangent to P₂ which passes through (f₁, 0). If m1 is the slope of T₁ and m₂ is the slope of T₂, then the value of

 is (JEE Adv. 2015)

Ans. (4)

Sol.  Ellipse :

⇒ a = 3, b = 

∴ f₁ = 2 and f₂ = –2 P₁ : y² = 8x and P₂ : y² = –16x

T₁ : y = m₁x  +

It passes through (–4, 0),

It passes through (2, 0)