Question

The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on the $x-$axis and passing through $(2,1)$, is

• A. $x^2+y^2-x=0$
• B. $4x^2+2y^2-9y=0$
• C. $2x^2+4y^2-9x=0$
• D. $4x^2+2y^2-9x=0$

Answer 1

The correct option is D $4x^2+2y^2-9x=0$

Equation of the normal at $(x_1,y_1)$ is
$y-y_1=\frac{dx}{dy}(x-x_1)$
Put $y=0$, then
$x=x_1+y_1\frac{dy}{dx}$
$y_1^2=2x_{1}x$
$=2x_1(x_1+y_1\frac{dy}{dx})$
$\Rightarrow y_1^2=2x_1^2+2x_1{y}_1\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{y_1^2-2x_1^2}{2x_1{y}_1}=\frac{{\left(\frac{y_1}{x_1}\right)}^2-2}{2\left(\frac{y_1}{x_1}\right)}$
$\Rightarrow \frac{dy}{dx}=\frac{{\left(\frac{y}{x}\right)}^2-2}{2y/x}$
Put $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$,
We have, $v+\frac{xdv}{dx}=\frac{v^2-2}{2v}$
$\Rightarrow x\cdot \frac{dv}{dx}=-\frac{2+v^2)}{2v}$
$\Rightarrow \frac{2v\cdot dv}{2+v^2}+\frac{dx}{x}=0$
On integrating both sides, we get
$\log (2+v^2)+\log |x|=\log C$
$\Rightarrow \log (2+v^2)|x|=\log C$
$\Rightarrow |x|(2+\frac{y^2}{x^2})=C$
Since it passes through $(2,1)$ then, $2(2+\frac{1}{4})=C$
Thus $C=\frac{9}{2}$
Therefore, $\frac{|x|(2x^2+y^2)}{x^2}=\frac{9}{2}$
$\Rightarrow 2x^2+y^2=\frac{9}{2}|x|$
$\Rightarrow 4x^2+2y^2-9|x|=0$.