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 $(2\sqrt{2}a,2\sqrt{2}a)$ is

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

Answer 1

Answer: (D)

$8a^2$

Point on the curve is $(2\sqrt{2}a,2\sqrt{2}a)$

Equation of tangent at this point is $x_{1}x+y_{1}y=16a^2$

here $x_1=2\sqrt{2}a$ and $y_1=2\sqrt{2}a$

$Area=\frac{1}{2}\times base\times height$

Point of intersection of tangent to the x axis is $\frac{16a^2}{2\sqrt{2}a}$

So,

Base = $4\sqrt{2}a$

Height =$2\sqrt{2}a$

$Area=\frac{1}{2}\times 4\sqrt{2}a\times 2\sqrt{2}a$

$Area=8a^2$