Question

Find the equation of the normal to the curve x² + 2y² − 4x − 6y + 8 = 0 at the point whose abscissa is 2.

Answer 1

Abscissa means the horizontal co-ordiante of a point.
Given that abscissa = 2.
i.e., x = 2

$$\begin{gathered} x^2+2y^2-4x-6y+8=0...\left(1\right) \\ \text{Differentiating both sides w.r.t.}x, \\ 2x+4y\frac{dy}{dx}-4-6\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}\left(4y-6\right)=4-2x \\ \Rightarrow \frac{dy}{dx}=\frac{4-2x}{4y-6}=\frac{2-x}{2y-3} \\ \text{When}\mathit{\text{x}}\text{=2, from (1), we get} \\ 4+2y^2-8-6y+8=0 \\ \Rightarrow 2y^2-6y+4=0 \\ \Rightarrow y^2-3y+2=0 \\ \Rightarrow \left(y-1\right)\left(y-2\right)=0 \\ \Rightarrow y=1\text{or}y=2 \\ \text{Case-1:}y=1 \\ \text{Slope of tangent =}{\left(\frac{dy}{dx}\right)}_{\left(2,1\right)}\text{=}\frac{0}{-1}\text{=0} \\ \left(x_1,y_1\right)=\left(2,1\right) \\ \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-1=\frac{-1}{0}\left(x-2\right) \\ \Rightarrow x-2=0 \\ \Rightarrow x=2 \\ \text{Case-2:}y=2 \\ \text{Slope of tangent =}{\left(\frac{dy}{dx}\right)}_{\left(2,2\right)}\text{=}\frac{0}{1}\text{=0} \\ \left(x_1,y_1\right)=\left(2,2\right) \\ \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-2=\frac{-1}{0}\left(x-2\right) \\ \Rightarrow x-2=0 \\ \Rightarrow x=2 \end{gathered}$$

In both cases, the equation of normal is x = 2