Find the coordinates of all the points P on the ellipse (x^2)/(a^2) + (y^2)/(b^2) = 1, for which the area of the Δ PON is maximum, where O denotes the origin and N is the foot of the perpendicular from O to the tangent at P.

(dfrac± a^2√(a^2+b^2), dfrac± b^2√(a^2+b^2) )

(dfrac± a^2√(a^2-b^2), dfrac± b^2√(a^2-b^2) )

(dfrac± a^2√(a^2+b^2), dfrac± b^2√(a^2+b^2) )

(dfrac± a^2√(a^2-b^2), dfrac± b^2√(a^2+b^2) )

Find the coordinates of all the points P on the ellipse… — JEE Main 1996 Mathematics PYQ

1996

Find the coordinates of all the points

P

on the ellipse

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

, for which the area of the

\Delta PON

is maximum, where

O

denotes the origin and

N

is the foot of the perpendicular from

O

to the tangent at

P

A.
$\left(\dfrac{\pm a^2}{\sqrt{a^2+b^2}}, \dfrac{\pm b^2}{\sqrt{a^2+b^2}}\right)$
B.
$\left(\dfrac{\pm a^2}{\sqrt{a^2-b^2}}, \dfrac{\pm b^2}{\sqrt{a^2-b^2}}\right)$
C.
$\left(\dfrac{\pm a^2}{\sqrt{a^2+b^2}}, \dfrac{\pm b^2}{\sqrt{a^2+b^2}}\right)$
D.
$\left(\dfrac{\pm a^2}{\sqrt{a^2-b^2}}, \dfrac{\pm b^2}{\sqrt{a^2+b^2}}\right)$

Show answer

Correct option: A

[1996]

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