Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced

The given curve is  (an ellipse) and given line is y = 4x + c.
We know that y = mx + c touches the ellipse

Here

∴ two values of c exist

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.

Let   be the tangent to 

where  

and pts of contact are  

or 

if y = mx + c is tangent to y = x² then x² – mx – c = 0 has equal roots

⇒ m² + 4c = 0 ⇒ c

is tangent to y = x²

∴ This is also tangent to  y = – (x – 2)²

= x² + 4x – 4

⇒ x² + (m – 4)x += 0

has equal roots⇒ m² – 8m + 16 = – m² + 16

⇒ m = 0, 4 ⇒ y = 0 or y = 4x – 4 are the tangents.

For the given ellipse

⇒ Eccentricity of hyperbola 

Let the hyperbola be  = 1 then

 = 1
 As it passes through focus of ellipse
i.e. (3, 0)

∴ we get A² = 9 ⇒ B² = 16

∴ Equation of hyper bola isfocus of hyperbola is (5, 0), vertex of hyperbola is (3, 0).

Given ellipse is  x² + 4y² = 4

a = 2, b = 1

As per question 

∴ PQ = 2

Now if PQ is the length of latus rectum to be found, then PQ = 4a =

 Also as PQ is horizontal, parabola with PQ as latus rectum can be upward parabola (with vertex at A) or down ward parabola (with vertex at A').
For upward parabola,

AR = a =  ∴ Coordinates of A =

So equation of upward parabola is given by

or  

For downward parabola A'R = a = 

∴ Coordinates of A' =  

So equation of downward parabola is given by

  or 

∴ the equation of required parabola is given by equation (1) or (2).

In ΔABC , given that

But in a triangle sin 

Applying componendo and dividendo, we get

⇒ a + b + c  = 3a or  b + c = 2a
i.e.  AC+ AB = constant        
(∵ Base BC = a is given to be constant) ⇒ A moves on an ellipse.

Let P(at² , 2at) be any point on the parabola y² = 4ax.

Then tangent to parabola at P is  

which meets the axis of parabola i.e x-axis at T (– at², 0).
Also normal to parabola at P is tx + y = 2at+ at³ which meets the axis of parabola at N (2a + at² , 0)
Let G (x, y) be the centriod of Δ PTN , then

Eliminating t from above, we get the locus of centriod G as

which is a parabola with vertex  directrix as

latus r ectum as   and
focus  as (a, 0).

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

Given parabola y² = 4x

Let  A(t₁² ,2t₁ ) and B(t₂² ,2t₂) Then centre of circle drawn with AB as diameter is

As circle touches x-axis

∴ r = | t₁ + t₂| ⇒ t₁ + t₂ = ±r

Also slope of  AB 

For x² + 4y² = 4        or  

As per question, 

focus of ellipse is (±, 0) As hyperbola passes through (±, 0)

∴ b = 1 and focus of hyperbola (± 2, 0)

∴ Equation of hyperbola is  

or     x² – 3y² = 3

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.

KEY CONCEPT : If slope of tangent is m, then equations of tangents to hyperbola

= are

 with the points of contact

∵ Tangent to hyperbola = is parallel to 2x – y = 1, therefore slope of tangent = 2

∴ Points of contact are  

i.e.

Let point P in first quadrant, lying on parabola y² = 2x

be Let Q be the point  Clearly a > 0.

∵ PQ is the diameter of circle through P, O, Q

∴ ∠POQ = 90°  = –1 ⇒ ab = –4

⇒ b is negative.
Also ar. ΔPOQ = 3

⇒

 (using ab = –4)

As a is positive and b is negative, we have a – b = 3 

(using ab = –4)

⇒ a² – 3a + 4 = 0 ⇒ a² – 2a – a + 4 = 0
⇒ (a – 2) (a – ) = 0  ⇒ a = 2 ,

∴ Point P can be  or 

i.e. (4, 2) or (1, )

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 ⇒  

H : x² – y² = 1  S : Circle with centre N(x₂, 0) Common tangent to H and S at P(x₁, y₁) is

xx₁ – yy₁ = 1   ⇒

Also radius of circle S with centre N(x₂, 0) through point of contact (x₁, y₁) is perpendicular to tangent

⇒ x₁ = x₂ – x₁ or x₂ = 2x₁

M is the point of intersection of tangent at P and x-axis

∵  Centroid of ΔPMN is (ℓ, m)

and y₁ = 3m

Using x₂ = 2x₁,

Also (x₁, y₁) lies on H,

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 =

Let point P on parabola y² = 4x be (t², 2t)

∵ PS is shortest distance, therefore PS should be the normal to parabola.

Equation of normal to y² = 4x at P (t², 2t) is y – 2t = – t(x – t²)
It passes through S(2, 8)
∴ 8 – 2t = – t(2 – t²) ⇒ t³ = 8 or t = 2
∴ P(4, 4)
Also slope of tangent to circle at

Equation of normal at t = 2 is 2x + y = 12
Clearly x-intercept = 6
SP = and SQ = r = 2
∴ Q divides SP in the ratio SP : PQ
Hence a, c , d are the correct options.