Question

The abscissa of the point on the curve $a{y}^2=x^3$, then normal at which cuts off equal intercepts from the axes is-

• A. $1$
• B. $\frac{4a}{3}$
• C. $3$
• D. $\frac{4a}{9}$

Answer 1

The correct option is D $\frac{4a}{9}$

$N:\frac{x}{9}+\frac{y}{9}=1$

N: $x+y=1$

$m_N=-1$

$m_T=1$

$a{y}^2=x63$

$\Rightarrow y^2=\frac{x^3}{a}$

Differentiating with respect to x,

$2y\frac{dy}{dx}=\frac{3^2}{a}$

$\frac{dy}{dx}=\frac{3x^2}{2ay}$

$m_T=\frac{dy}{dx}=\frac{3x^2}{2ay}=1$

$\Rightarrow 3x^2=2ay$

Let $P(\alpha ,\beta )$ is a point lying on curve($\alpha {\beta }^2={\alpha }^3$)

$3{\alpha }^2=2a\beta$

$\alpha \beta =\frac{3{\alpha }^2}{2a}$

Putting value of $\beta$ in $\alpha {\beta }^2={\alpha }^3$

$\Rightarrow \frac{9{\alpha }^4}{4a^2}={\alpha }^3$

$\Rightarrow a=\frac{4a}{9}$