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

The correct option is A

$y=3x-5$

Explanation for the correct answer:

Finding the equation of the line,

Given equation $x^2+y^2-2x+4y-5=0$

By differentiating the equation we get

$$\begin{gathered} \frac{d}{dx}\left(x^2+y^2-2x+4y-5\right)=0 \\ \Rightarrow 2x+2y.\frac{dy}{dx}-2+4\frac{dy}{dx}=0 \\ \Rightarrow (2y+4)\frac{dy}{dx}=2-2x \\ \Rightarrow \frac{dy}{dx}=\frac{1-x}{y+2} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,1)}=\frac{1-2}{1+2} \\ \Rightarrow -{\left(\frac{dx}{dy}\right)}_{(2,1)}=3 \end{gathered}$$

Now, the equation of the normal is

$$\begin{gathered} y-y_1=\text{slope}(x-x_1) \\ \Rightarrow y-1=3(x-2) \\ \Rightarrow y-1=3x-6 \\ \Rightarrow y=3x-5 \end{gathered}$$

Hence, the correct answer is Option (A).