Question

The line normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4) then its equation is

• A. $x^2+y^2+6x-7=0$
• B. $2(x^2+y^2)-12x-7=0$
• C. $x^2+y^2-6x-7=0$
• D. None of these

Answer 1

Answer: (C)

The equation of normal to a curve at a point (x, y) is
$\left(Y-y\right)\frac{dy}{dx}+\left(X-x\right)=0$
Since it passes throgh the point $\left(3,0\right)$, we have
$\left(0-y\right)\frac{dy}{dx}+\left(3-x\right)=0\Rightarrow y\frac{dy}{dx}=\left(3-x\right)\Rightarrow ydy=\left(3-x\right)dx$
Integrating, we get $\frac{y^2}{2}=3x-\frac{x^2}{2}+C$
$\Rightarrow x^2+y^2-6x-2c=0$
Since the curve passes through $\left(3,4\right)$, we have
$9+16-18-2c=0\Rightarrow c=\frac{7}{2}$.
$\therefore x^2+y^2-6x-7=0$ is the required equation of the curve.