Question

The equation of the curve for which the square of the $y$ intercept by any tangent at any point $P(x,y)(xy>0)$ is equal to the product of the co-ordinates of the point of tangency is

• A. $\log (c+x)=\pm \sqrt{\frac{y}{x}}$
• B. $x=e^{\pm \sqrt{\frac{y}{x}}}+c$
• C. $cx=e^{\pm \sqrt{\frac{y}{x}}}$
• D. $\sqrt{cx}=e^{\pm \sqrt{\frac{y}{x}}}$

Answer 1

The correct option is D $\sqrt{cx}=e^{\pm \sqrt{\frac{y}{x}}}$

Equation of the tangent at any point,
$y=mx+c$
It passes through $P(x,y)$,
$c=y-x\frac{dy}{dx}$
According to the question,
${(y-x\frac{dy}{dx})}^2=xy$
$\Rightarrow y-x\frac{dy}{dx}=\pm \sqrt{xy}$
$\frac{dy}{dx}=\frac{y\pm \sqrt{xy}}{x}$

Let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=v\pm \sqrt{v}$
$\Rightarrow \pm \int \frac{dv}{\sqrt{v}}=\int \frac{dx}{x}$
$\Rightarrow \pm 2\sqrt{v}=\log x+\log c$
$\Rightarrow \pm \sqrt{\frac{y}{x}}=\log \sqrt{cx}$
$\Rightarrow \sqrt{cx}=e^{\pm \sqrt{\frac{y}{x}}}$