Question

The equation of the normal of ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-5=0$ at $\left(2,1\right)$ 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 options:

Step 1: Finding the value of $m_1$ and $m_2$

Given equation of curve is ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-5=0$

On differentiating the above equation we get,

$\begin{array}{rcl}2\mathrm{x}+2\mathrm{y}\frac{d\mathrm{y}}{d\mathrm{x}}-2+4\frac{d\mathrm{y}}{d\mathrm{x}}& =& 0\\ \frac{d\mathrm{y}}{d\mathrm{x}}\left(2y+4\right)& =& 2-2x\\ \frac{d\mathrm{y}}{d\mathrm{x}}& =& \frac{2-2x}{2y+4}\\ {\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)}_{2,1}& =& \frac{2-2\left(2\right)}{2\left(1\right)+4}\\ {\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)}_{2,1}& =& \frac{-2}{6}\\ {\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)}_{2,1}& =& \frac{-1}{3}\end{array}$

Thus, $m_1=\frac{-1}{3}$

We know that $m_1.m_2=-1$. Substituting the value of $m_1$, we get

$$\begin{gathered} \left(-\frac{1}{3}\right).m_2=-1 \\ {m}_2=3 \end{gathered}$$

Step 2: Finding the equation of normal

So, the equation of normal at point $\left(2,1\right)$ is

$\begin{array}{rcl}y-y_1& =& m\left(x-x_1\right)\\ y-1& =& 3\left(x-2\right)\\ y-1& =& 3x-6\\ y& =& 3x-6+1\\ y& =& 3x-5\end{array}$

Thus, the equation of the normal of ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-5=0$ at $\left(2,1\right)$ is $\mathrm{y}=3\mathrm{x}-5$

Hence, option (A) is the correct answer.