JEE Advanced 2022 Paper 2 · Mathematics · Conic Sections · medium

The parabola y^2=4x and the ellipse (x^2/(4)+(y^2/(3)=1 have exactly two common tangents. These two tangents together with the y-axis form a…

Numerical answer

The parabola

y^{2}=4x

and the ellipse

\dfrac{x^{2}}{4}+\dfrac{y^{2}}{3}=1

have exactly two common tangents. These two tangents together with the

y

-axis form a triangle. The area of this triangle, in square units, is

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Answer: 8

A tangent to

y^2=4x

with slope

m

y=mx+\frac1m

. It touches the ellipse

\frac{x^2}{4}+\frac{y^2}{3}=1

when

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

, i.e.

\frac1{m^2}=4m^2+3

. Setting

u=m^2

gives

4u^2+3u-1=0

, so

(4u-1)(u+1)=0

and

u=\tfrac14

(the root

u=-1

is rejected). Thus

m=\pm\tfrac12

and the common tangents are

y=\tfrac x2+2

and

y=-\tfrac x2-2

. They meet where

\tfrac x2+2=-\tfrac x2-2

, i.e. at

(-4,0)

, and cut the

y

-axis at

(0,2)

and

(0,-2)

. The triangle has base

4

along the

y

-axis and height

4

, so its area is

\tfrac12\cdot4\cdot4=8

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