Let $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ be two distinct common tangents to the ellipse $\mathrm{E}: \frac{\mathrm{x}^{2}}{6}+\frac{\mathrm{y}^{2}}{3}=1$ and the parabola $\mathrm{P}: \mathrm{y}^{2}=12 \mathrm{x}$. Suppose that the tangent $\mathrm{T}_{1}$ touches P and E at the point $\mathrm{A}_{1}$ and $\mathrm{A}_{2}$, respectively and the tangent $\mathrm{T}_{2}$ touches P and E at the points $\mathrm{A}_{4}$ and $\mathrm{A}_{3}$, respectively. Then which of the following statements is(are) true?

(A) The area of the quadrilateral $\mathrm{A}_{1} \mathrm{~A}_{2} \mathrm{~A}_{3} \mathrm{~A}_{4}$ is 35 square units; (C) The tangents $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ meet the x -axis at the point $(-3,0)$

(B) The area of the quadrilateral $\mathrm{A}_{1} \mathrm{~A}_{2} \mathrm{~A}_{3} \mathrm{~A}_{4}$ is 36 square units

(D) The tangents $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ meet the x-axis at the point (-6, 0)

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Parabola

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Let

T_{1}

and

T_{2}

be two distinct common tangents to the ellipse

E : \frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1

and the parabola

P : y^{2} = 12 x

. Suppose that the tangent

T_{1}

touches P and E at the point

A_{1}

and

A_{2}

, respectively and the tangent

T_{2}

touches P and E at the points

A_{4}

and

A_{3}

, respectively. Then which of the following statements is(are) true?

(A)The area of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is 35 square units✓ Correct
(B)The area of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is 36 square units
(C)The tangents $T_{1}$ and $T_{2}$ meet the x -axis at the point $(- 3, 0)$ ✓ Correct
(D)The tangents $T_{1}$ and $T_{2}$ meet the x-axis at the point (-6, 0)

Answer: (A), (C)

Solution

The tangent to parabola

y^{2} = 12 x

y = m x + \frac{3}{m}

.

For this to be tangent to ellipse

\frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1

, we need:

c^{2} = a^{2} m^{2} + b^{2}

\frac{9}{m ^{2}} = 6 m^{2} + 3

m^{2} = 1 \Rightarrow m = \pm 1

The two common tangents are

y = x + 3

and

y = - x - 3

. Both cut the x-axis at

(- 3, 0)

.

For

y = x + 3

A_{1} (3, 6)

on parabola,

A_{2} (- 2, 1)

on ellipse.

For

y = - x - 3

A_{4} (3, - 6)

on parabola,

A_{3} (- 2, - 1)

on ellipse.

The quadrilateral has parallel sides

A_{1} A_{4} = 12

and

A_{2} A_{3} = 2

, with distance between them = 5.

Area

= \frac{1}{2} (12 + 2) \times 5 = 35

square units.

Answer: (A), (C)

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Parabola

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