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Q.
A normal is drawn at a point $P(x, y)$ of a curve, It meets the $x$-axis at $Q$. If $P Q$ is of constant length $k$. Such a curve passing through $(0, k)$ is :
Differential Equations
Solution:
Equation of normal at $P ( x , y )$ is
$Y-y=-\frac{d x}{d y}(X-x)$
so coordinate of $Q$ is $\left(x+y \frac{d y}{d x}, 0\right)$
Thus $(P Q)^2=(X-x)^2+(0-y)^2$
$\Rightarrow k^2=\left(y \frac{d y}{d x}\right)^2+y^2$
$ \Rightarrow y \frac{d y}{d x}=\pm \sqrt{k^2-y^2} $
$\Rightarrow \frac{y d y}{\sqrt{k^2-y^2}}=\pm \int d x $
$\Rightarrow =\pm x+c $
$\Rightarrow $ it passes through $(0, k)$
$\Rightarrow c=0 \Rightarrow -\sqrt{k^2-y^2}=\pm x$
$k^2-y^2=x^2 $
$x^2+y^2=k^2$