JEE Main 2022 · Mathematics · Conic Sections · medium

Let P be the point in the first quadrant on the ellipse (x^2/(16)+(y^2/(12)=1 whose abscissa is 2. The tangent at P meets the coordinate axes at A…

Single correct option

Let

P

be the point in the first quadrant on the ellipse

\dfrac{x^2}{16}+\dfrac{y^2}{12}=1

whose abscissa is

2

. The tangent at

P

meets the coordinate axes at

A

and

B

. The area of triangle

OAB

, where

O

is the origin, is

Show answer and solution

Answer: (B) $16$

Put

x=2

\frac{y^2}{12}=1-\frac4{16}=\frac34

, so

y^2=9

and

P=(2,3)

in the first quadrant. The tangent at

(x_1,y_1)

\frac{xx_1}{16}+\frac{yy_1}{12}=1

, i.e.

\frac{2x}{16}+\frac{3y}{12}=1

, which simplifies to

\frac x8+\frac y4=1

. It meets the axes at

A(8,0)

and

B(0,4)

. The area of right triangle

OAB

\frac12\cdot8\cdot4=16

. The product

OA\cdot OB=32

is the trap answer.

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