Tangent at a point on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
is drawn which cuts the coordinates axes at A and B the maximum area of the triangle OAB is ( O being origin )

Options

A. ab
B. `(a^(3)+b^(3)+ab)/(3)`
C. `a^(2)+b^(2)`
D. `(a^(2)+b^(2))/(4)`

Answer

A

Explanation

Let P `(a cos theta , bsin theta )` be any point on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`, the equation of tangent to the ellipse at P is
`(x)/(a) cos theta +(y)/(b) sin theta=1`
this cuts x-axis ata `A((a)/( cos theta ,0) ) and " y- axes at " (0,(b) /( sin theta))`
` therefore Delta = "Area of "Delta OAB =(1)/(2) (OAxxOB) =(ab)/( 2 cos theta sin theta )= ( ab)/( sin 2 theta)`
Clearly ,` Delta ` is minimum when sin `2 theta =1i.e., theta =(pi)/(4).`
minimum value of `Delta ` is ab.