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 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

$4x^2+2y^2–9x=0$

Explanation for the correct option:

Finding the equation of the line passing through the given point:

Step 1: Finding general equation

The general equation of normal at the point $\left(x_1,y_1\right)$ is $y-y_1=-\frac{dx}{dy}\left(x-x_1\right)$

Substituting $y=0$ we get $x=x_1+y_1$

Now, substitute

$$\begin{gathered} y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \mathrm{Now}, \\ v+x\frac{dv}{dx}=\frac{v^2{x}^2-2x^2}{2xvx} \\ =\frac{v^2-2}{2v} \end{gathered}$$

Now, solve it more we get,

$$\begin{gathered} x\frac{\mathit{d}v}{\mathit{d}x}\mathit{=}\frac{v^{\mathit{2}}\mathit{-}\mathit{2}}{\mathit{2}v}\mathit{-}v \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{v^{\mathit{2}}\mathit{-}\mathit{2}\mathit{-}\mathit{2}{v}^{\mathit{2}}}{\mathit{2}v} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{\mathit{-}{v}^{\mathit{2}}\mathit{-}\mathit{2}}{\mathit{2}v} \end{gathered}$$

Step 2: Finding value of k

Now integrating,

$$\begin{gathered} \int \frac{2vdv}{v^2+2}=-\int \frac{dx}{x} \\ \mathrm{Let}{v}^2+2=t\mathrm{and}2vdv=dt \\ \Rightarrow \int \frac{dt}{d}=-\ln \left(x\right)+\log \left(k\right) \\ \Rightarrow \ln \left(v^2+2\right)+\ln \left(x\right)=\ln \left(k\right) \\ \Rightarrow \ln \left(\frac{y^2}{x^2}+2\right)+\ln \left(x\right)=\ln \left(k\right) \\ \Rightarrow \ln \left(\frac{y^2}{x^2}+2x\right)=\ln \left(k\right) \\ \Rightarrow y^2+2x^2=kx\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right) \end{gathered}$$

Since it is passing through $\left(2,1\right)$

$$\begin{gathered} 1+8-2k=0 \\ \Rightarrow 9-2k=0 \\ \Rightarrow k=\frac{9}{2} \end{gathered}$$

So putting $\mathrm{k}=\frac{9}{2}$ in $\left(1\right)$

$$\begin{gathered} \Rightarrow 2x^2+y^2-\frac{9}{2}x=0 \\ \Rightarrow 4x^2+2y^2-9x=0 \end{gathered}$$

Therefore, the correct answer is option (D).