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. Such a curve passing through (0, k) is:

• A. a circle with curve (0, 0)
• B. $x^2+y^2=k^2$
• C. $(1+k)x^2+y^2=k^2$
• D. $x^2+(1+k^2)y^2=k^2$

Answer 1

Answer: (A), (B)

The correct options are
A a circle with curve (0, 0)
B $x^2+y^2=k^2$

The equation of normal at P(x, y) is
$Y-y=-\frac{dx}{dy}(X-x)$
So the coordinates of Q are $(x+y\frac{dy}{dx},0)$
Thus $P{Q}^2=(X-x{)}^2+(0-y{)}^2={\left(y\frac{dy}{dx}\right)}^2+y^2$
$\Rightarrow k^2={\left(y\frac{dy}{dx}\right)}^2+y^2$
$\Rightarrow {\left(y\frac{dy}{dx}\right)}^2=k^2-y^2\Rightarrow y\frac{dy}{dx}=\pm \sqrt{k^2-y^2}$
To find the equation of the curve we rewrite it as
$\frac{ydy}{\sqrt{k^2-y^2}}=\pm dx$.
Integrating, we get
$\int \frac{ydy}{\sqrt{k^2-y^2}}=\pm \int dx$
$\Rightarrow -\sqrt{k^2-y^2}=\pm x+c$ (3)
As the curve passes through (0, k) we get $-\sqrt{k^2-k^2}=\pm \left(0\right)+c\Rightarrow =0$
Therefore, (3) can be written as $-\sqrt{k^2-y^2}=\pm x\Rightarrow k^2-y^2=x^2$ or $x^2+y^2=k^2$.