Question

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

• A. $x^2+y^2-x=0$
• B. $4x^2+2y^2-9y=0$
• C. $2x^2+4y^{2}9-x=0$
• 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 $(x,y)$ is $Y-y=-\frac{dx}{dy}(X-x)$
$\therefore$ x-intercept $y\frac{dx}{dy}+x$ (Putting $Y=0$)
Given: $\Rightarrow \frac{dy}{dx}=\frac{y^2-{2x}^2}{2xy}.$ Put $y-vx.$ Then $v+x\frac{dy}{dx}=\frac{v^2-2}{2v}\Rightarrow x\frac{dv}{dx}=\frac{-(2+v^2)}{2v}$
$\Rightarrow \int \frac{2v}{v^2+2}dv+\int \frac{dx}{x}=\log k\Rightarrow \log (v^2+2)=\log x=\log k$
$\Rightarrow x(v^2+2)=k\Rightarrow y^2+{2x}^2-kx=0$
If this passes through$(2,1),i=\frac{9}{2}$
Then, the equation becomes ${4x}^2+{2y}^3-9x=0.$