Paragraph: Let F_ 1 (x_ 1 , 0) and F_ 2 (x_ 2 , 0) , for x_ 1 0 , be the foci of the ellipse x^ 2 9 + y^ 2 8 =1 . Suppose a parabola having vertex at the origin and focus at F_ 2 intersects the ellipse at point M in the

Correct answer: C. 5: 8 Equation of chord of contact MN is x = (3)/(2) Let R(h,k) ⇒ Equation of chord of contact ⇒ (xh)/(9) + (yk)/(8) = 1 Comparing we get R(6,0) Normal to parabola at M: y - 6 = - 6 2 · 1 (x - (3)/(2)) Solving it with y = 0, we get Q ≡ ((7)/(2), 0) ∴ Area of triangle MQR = (1)/(2) . (6 - (7)/(2)) . 6 = 5 6 4 Area of quadrilateral M F _ 1 N F _ 2 = 2 · (1)/(2)(1 - (- 1)) 6 = 2 6 Required ratio = 5: 8

JEE Advanced 2016Paper 2

Exam: JEE Advanced
Year: 2016
Paper: Paper 2
Subject: Mathematics
Chapter: Ellipse

JEE Advanced 2016 · Paper 2 · Mathematics · Ellipse

Paragraph: Let F_ 1 (x_ 1 , 0) and F_ 2 (x_ 2 , 0) , for x_ 1 0 ,...

Paragraph: Let

F_{1}

\left(x_{1}, 0\right)

and

F_{2}\left(x_{2}, 0\right)

, for

x_{1}<0

and

x_{2}>0

, be the foci of the ellipse

\frac{x^{2}}{9}+\frac{y^{2}}{8}=1 .

Suppose a parabola having vertex at the origin and focus at

F_{2}

intersects the ellipse at point

in the first quadrant and at point

in the fourth quadrant. <br> <strong> Question: </strong> If the tangents to the ellipse at

and

meet at

and the normal to the parabola at

meets the

-axis at

, then the ratio of area of the triangle

M Q R

to area of the quadrilateral

M F_{1} N F_{2} is Paragraph: Let F_ 1 (x_ 1 , 0) and F_ 2 (x_ 2 , 0) , for x_ 1 0 , be the foci of the ellipse x^ 2 9 + y^ 2 8 =1 .

Options

A $3: 4$
B $4: 5$
C $5: 8$
D $2: 3$

Correct answer

C. $5: 8$

Step-by-step solution

Equation of chord of contact MN is

x = \dfrac{3}{2}

Let R(h,k)

⇒

Equation of chord of contact

⇒ \dfrac{xh}{9} + \dfrac{yk}{8} = 1

Comparing we get R(6,0)
Normal to parabola at

M: y - \sqrt{6} = - \dfrac{\sqrt{6}}{2 \cdot 1}\left(x - \dfrac{3}{2}\right)

Solving it with

y = 0,

we get

Q ≡ \left(\dfrac{7}{2}, 0\right)

∴

Area of triangle MQR

= \dfrac{1}{2} . \left(6 - \dfrac{7}{2}\right) . \sqrt{6} = \dfrac{5\sqrt{6}}{4}

Area of quadrilateral

M{F}_{1}N{F}_{2} = 2 \cdot \dfrac{1}{2}\left(1 - \left(- 1\right)\right)\sqrt{6} = 2\sqrt{6}

Required ratio

= 5: 8

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