Question

The area of the triangle formed by the positive x-axis, the tangent and normal to the curve $x^2+y^2=16a^2$ at the point $(2\sqrt{2a},2\sqrt{2a})$ is

• A. $a^2$
• B. $16a^2$
• C. $4a^2$
• D. $8a^2$

Answer 1

The correct option is D $8a^2$

Radius $=\sqrt{16a^2}$ $=4a$

As the normal is $x=y$, the triangle is isosceles, right angled, so $PT=PO=4a$

So, area $=\frac{1}{2}\times 4a\times 4a$

$=8a^2$