P is a variable point on the ellipse with foci `S_1` and `S_2`. If A is the area of the the triangle `PS_1S_2`, the maximum value of A is

Options

A. ab
B. abe
C. `(1)/(2) ab`
D. `(1)/(2) abe`

Answer

B

Explanation

Let P(a cos `theta ` ,b sin `theta` ) be the variable point on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =1` , then
A = Area of `Delta PS_(1) S_(2)`
`implies A=(1)/(2)|{:(a cos theta, b sin theta,1),( ae,0,1),(-ae ,0,1):}|=(1)/(2) b sin theta xx2ae = abc sin theta` Clearly it is maximum when `theta =(pi)/(2)` and the maximum value of A is able
ALITER Let P( a cos ` theta , b sin theta) ` be a variable point on the ellipse
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
then ,
`A =Area " of" Delta Ps_(1) S_(2)`
`implies A=(1)/(2) (" Base "xx""Height ")`
` implies A=(1)/(2) (S_(1) S_(2) xxPM),` where PM is the foot of the perpendicular from P. on major axis
`implies A=(1)/(2) ( 2aexx b sin theta)=abe sin theta`
Clearly ,A is maximum when ` theta =pi//2` and the maximum value is A is abc.