$\Rightarrow 2(x-h)+2(y-k) \frac{d y}{d x}=0$
$\Rightarrow(x-h)+(y-k) \frac{d y}{d x}=0 \quad \ldots$ (ii)
Again differentiating $(y-k)=-\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$
Putting in Eq. (ii), we get $x-h=-(y-k) \frac{d y}{d x}$
$=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right] \frac{d y}{d x}}{\frac{d^{2} y}{d x^{2}}}$
Putting in Eq. (i), we get $=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left(\frac{d y}{d x}\right)^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}+\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}=a^{2}$
$\Rightarrow\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left[\left(\frac{d y}{d x}\right)^{2}+1\right]=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$
$\Rightarrow\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$
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{where $x,y \in R^+, x^2y + x \ne 0$ }