The area of the smaller region enclosed by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1, $ and the straight line $\frac{x}{4}+\frac{y}{3}=1$ is $a\pi + b $. The value of $a+b $ is :

-3

The correct answer is Option (1) → -3

area required = area of quarter ellipse - area of triangle

$\frac{πab}{4}-\frac{1}{2}ab$

$=\frac{π(4)(3)}{4}-\frac{1}{2}(4)(3)=aπ+b$

$3π-6=aπ+b$

$a=3,b=-6$

$a+b=-3$