Question

The area bounded by the tangent and normal to the curve $y(6-x)=x^2$ at $(3,3)$ and the x-axis is

• A. $5$
• B. $6$
• C. $15$
• D. $3$

Answer 1

The correct option is C $15$

$y(6-x)=x^2$

$\Rightarrow 6\frac{dy}{dx}-(x\frac{dy}{dx}+y)=2x$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(3,3)}=3$

Hence the equations of the tangent and normal at these points are

$y-3=3(x-3)$ and $y-3=-\frac{1}{3}(x-3)$ repectively

These cut the x- axis at $(2,0)$ and $(12,0)$

$\therefore$ The area of the triangle is $\frac{1}{2}\left[3\right(0-0)+2(0-3)+12(3-0\left)\right]=\frac{1}{2}\left[30\right]=15squnits.$