Let T₁ and T₂ be two distinct common tangents to the ellipse E: \;x²/6+\;y²/3=1 and the parabola P: y²=12 x. Suppose that the tangent T₁ touches P and E at the points A₁ and A₂, respectively and the tangent T₂ touches P and E at the points A₄ and A₃, respectively. Then which of the following statements is(are) true? [JEE Adv 2023 P1]

The area of the quadrilateral A₁ A₂ A₃ A₄ is 35 square units

The area of the quadrilateral A₁ A₂ A₃ A₄ is 36 square units

The tangents T₁ and T₂ meet the x-axis at the point (-3,0)

The tangents T₁ and T₂ meet the x-axis at the point (-6,0)

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Ellipse

Section: Mathematics

Question : 3 of 25

Marks: +1, -0

Let

T_1

T_2

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

P: y^2=12 x

T_1

P

E

A_1

A_2

T_2

P

E

A_4

A_3

Solution:

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

, Tangent

: y=m_1 x \pm \sqrt{6 m_1^2+3}

P: y^2=12 x

, Tangent:

y=m_2 x+\;\frac{3}{m_2}

For common tangent

m=m_1=m_2, \pm \sqrt{6 m_1^2+3}=\;\frac{3}{m_2}

\Rightarrow m= \pm 1

\Rightarrow

equation of common tangents

y=x+3

and

y=-x-3

point of contact for parabola is

\left(\;\frac{a}{m^2}, \;\frac{2a}{m}\right)

\Rightarrow A_1 \equiv (3,6), \;\; A_4(3-6)

Let

A_2 (x_1, y_1) \Rightarrow

tangent to

E

\;\frac{x x_1}{6}+\;\frac{y y_1}{3}=1

A_3

is mirror image of

A_2

x

-axis

\Rightarrow A_3(-2,-1)

Intersection point of

T_1=0

and

T_2=0

(-3,0)

Area of quadrilateral

A_1 A_2 A_3 A_4=\;\frac{1}{2}(12+2) \times 5=35

square units

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