Question

The equation of normal of $x^2 + y^2 - 2x + 4y - 5 = 0$ at $(2, 1)$ is

• A. $y = 3x - 5$
• B. $2y = 3x - 4$
• C. $y = 3x + 4$
• D. $y = x + 1$

Answer 1

Answer: (A)

Given equation is
$x^{2}+y^{2}-2 x+4 y-5=0$
On differentiating, we get
$2 x+2 y \frac{d y}{d x}-2+4 \frac{d y}{d x}=0$
$\Rightarrow (y+2) \frac{d y}{d x}=1-x $
$\Rightarrow \left(\frac{d y}{d x}\right)_{(2,1)}=\frac{1-2}{1+2}=\frac{-1}{3}$
$\Rightarrow -\left(\frac{d x}{d y}\right)_{(21)}=3$
Now, equation of normal is
$(y-1) =3(x-2) $
$y-1 =3 x-6 $
$\Rightarrow y=3 x-5$