Question

The area of the triangle formed by the tan-gent and the normal at $(a,a)$ on the curve $y^2=\frac{x^3}{2a-x}$ and the line $x=2a$ is (in sq.units)

• A. $\frac{7a^2}{4}$
• B. $\frac{a^2}{2}$
• C. $\frac{5a^2}{4}$
• D. $\frac{9a^2}{4}$

Answer 1

The correct option is C $\frac{5a^2}{4}$

Equation of the given curve is $y^2(2a-x)=x^3$

Differentiating both sides w.r.t $x,$ we get

$y^2\times (-1)+(2a-x)\times 2y\frac{dy}{dx}=3x^2$

$\Rightarrow (2a-x)2y\frac{dy}{dx}=3x^2+y^2$

$\Rightarrow \frac{dy}{dx}=\frac{3x^2+y^2}{2y(2a-x)}$

Slope of tangent at $(a,a)$ is $m_1=\frac{3a^2+a^2}{2a(2a-a)}=\frac{4a^2}{2a^2}=2$

$\therefore$Slope of normal at $(a,a)$ is $m_2=-\frac{1}{2}$

Equation of tangent at $(a,a)$ is

$y-a=2(x-a)$

$\therefore y=2x-2a+a=2x-a$ ......$\left(1\right)$

Equation of normal at $(a,a)$ is

$y-a=-\frac{1}{2}(x-a)$

$\therefore 2y=-x+a+2a=-x+3a$

$\Rightarrow y=\frac{-x}{2}+\frac{3a}{2}$ ......$\left(2\right)$

Equation of the given line is

$x=2a$ ...$\left(3\right)$

Solving $\left(1\right)$ and $\left(3\right)$, we have

$x=2a$ and $y=2\left(2a\right)-a=4a–a=3a$

Solving $\left(2\right)$ and $\left(3\right)$, we have

$x=2a$ and $y=-\frac{2a}{2}+\frac{3a}{2}=\frac{a}{2}$

Solving $\left(1\right)$ and $\left(2\right)$, we have

$x=a$ and $y=a$

Let $A(2a,3a),B(2a,\frac{a}{2})$ and $C(a,a)$ be the vertices of the triangle.

Area of $△ABC=\frac{1}{2}|2a(\frac{a}{2}-a)+2a(a-3a)+a(3a-\frac{a}{2})|$

$=\frac{1}{2}|2a\left(\frac{-a}{2}\right)+2a(-2a)+a\left(\frac{5a}{2}\right)|$

$=\frac{1}{2}|-a^2-4a^2+\frac{5a^2}{2}|$

$=\frac{1}{2}\left|\frac{-10a^2+5a^2}{2}\right|$

$=\frac{5a^2}{4}$

Thus, the area of triangle formed is $\frac{5a^2}{4}$ square units.