Mathematics - Application of Derivative Question with Solution | TestHub

Equation of the normal at a point $P(x,y)$ is given by $Y-y=-\frac{1}{dy/dx}(X-x)$ Let the point Q at the x -axis be $\left(x_1,0\right)$. From (1) we get $y\frac{dy}{dx}=x_1-x$

Now given that $P{Q}^2=k^2$. We have ${\left(x-x_1\right)}^2+y^2=k^2$ or $x-x_1=\pm \sqrt{k^2-y^2}$ Hence using (2) we obtain $y\frac{dy}{dx}=\pm \sqrt{k^2-y^2}$ (3) is the required differential equation for such curves Now solving (3) we get $\int \frac{-ydy}{\sqrt{k^2-y^2}}=\int -dx$ or $x^2+y^2=k^2$, which is a circle with centre as origin and passes through both $(0,k)$ and $(k,0)$