Question

A normal is drawn at a point P (x, y) of a curve. It meets the x-axis and the y-axis in point A and B, respectively, such that $\frac{1}{OA}$ + $\frac{1}{OB}$ = 1, where O is the origin, find the equation of such a curve passing through (5, 4) –

• A. (x – 1)² + (y – 1)² = 16
• B. (x – 1)² + (y – 1)² = 25
• C. (x – 2)² + (y – 2)² = 9
• D. None of these

Answer 1

Answer: (B)

(x – 1)² + (y – 1)² = 25

Answer 2

The equation of the normal at (x, y) is :

(X – x) + (Y – y) $\frac{dy}{dx}$ = 0

Ž $\frac{X}{x + y\frac{dy}{dx}}$+ $\frac{Y}{\frac{(x + ydy/dx)}{dy/dx}}$ = 1

Ž OA = x + y $\frac{dy}{dx}$, OB = $\frac{\left( x + y\frac{dy}{dx} \right)}{\frac{dy}{dx}}$

Also, $\frac{1}{OA} + \frac{1}{OB}$ = 1

Ž 1 + $\frac{dy}{dx}$ = x + y$\frac{dy}{dx}$ Ž (y – 1) $\frac{dy}{dx}$ + (x – 1) = 0

Integrating, we get

(y – 1)² + (x – 1)² = c

Since the curve passes through (5, 4), c = 25.

Hence, the curve is (x – 1)² + (y – 1)² = 25