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 $\frac{dy}{dx}$ = ± $\sqrt{k^{2} - y^{2}}$ and the equation of such a curve passing through (0, k) –

• A. x² + y² = k
• B. x² + y² = k²
• C. x² – y² = k²
• D. None of these

Answer 1

Answer: (B)

x² + y² = k²

Answer 2

Let y = ƒ(x) be the curve such that the normal at P (x, y) to this curve meets x-axis at Q. Then,

PQ = length of the normal at P.

= y $\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}$

But PQ = k

\ y $\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}$ = k

Ž y² + y² $\left( \frac{dy}{dx} \right)^{2}$ = k²

Or y $\frac{dy}{dx}$ = ± $\sqrt{k^{2} - y^{2}}$, Integrating both sides, we get

$\frac{ydy}{\sqrt{k^{2} - y^{2}}} = \pm dx$,

–$\sqrt{k^{2} - y^{2}}$ = ± x + c, since it passes through (0, k) ® c = 0. \ – $\sqrt{k^{2} - y^{2}}$ = ± x

or k² – y² = x²

Ž x² + y² = k², is required equation of the curve