Question

A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k, then show that the differential equation describing such curves Is, $y\frac{dy}{dx}=\pm \sqrt{k^2-y^2}$. Find the equation of such a curve passing through (0.k), the curve is a:

• A. line
• B. circle
• C. parabola
• D. ellipse

Answer 1

The correct option is B circle

Equation of normal at point $P(x,y)$ to the curve is
$Y-y=-\frac{dx}{dy}(X-x)$
This straight line meets the x-axis in $O(X,0)$ where the number $X$ satisfies.
$0-y=-\frac{dx}{dy}(X-x)\Rightarrow X=x+y\frac{dy}{dx}$
Thus ${PO}^2={(X-x)}^2+{(0-y)}^2\Rightarrow {\left(y\frac{dy}{dx}\right)}^2$
$\Rightarrow y\frac{dy}{dx}=\pm \sqrt{k^2-y^2}=k^2-y^2$
To find the equation we rewrite it as $\frac{ydy}{\sqrt{k^2-y^2}}=\pm dx$
Integrating we get
$\int \frac{ydy}{\sqrt{k^2-y^2}}=\pm \int dx$
$\Rightarrow -\sqrt{k^2-y^2}=\pm x+c$ ...(1)
As the curve passes through $(0,k)$ we get
$-\sqrt{k^2-k^2}=\pm \left(o\right)+c\Rightarrow c=o$
Therefore equations (1) can be witten as $-\sqrt{k^2-y^2}=\pm x.$
$\Rightarrow k^2-y^2=x^2\Rightarrow x^2+y^2=k^2$