Question

Q. Find the equation of the normal at a point on the curve $x^2$=4y which passes through the point (1,2). also find the equation of the corresponding tangent.

Answer 1

Dear student
$$\begin{gathered} Theequationofthegivencurveis{x}^2=4y...\left(1\right) \\ Differentiating\left(1\right)w.r.tx,weget \\ 2x=4\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{2x}{4} \\ \Rightarrow \frac{dy}{dx}=\frac{x}{2} \\ Now,{\left(\frac{dy}{dx}\right)}_{\left(1,2\right)}=\frac{1}{2} \\ So,theequationof\tan gentat\left(1,2\right)isgivenby: \\ y-y_1={\left(\frac{dy}{dx}\right)}_{\left(1,2\right)}\left(x-x_1\right) \\ \Rightarrow y-2=\frac{1}{2}\left(x-1\right) \\ \Rightarrow 2y-4=x-1 \\ \Rightarrow 2y-x-4+1=0 \\ \Rightarrow 2y-x-3=0 \\ Equationofnormalat\left(1,2\right)isgivenby: \\ y-y_1=\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\left(1,2\right)}}\left(x-x_1\right) \\ \Rightarrow y-2=\frac{-1}{{\displaystyle \frac{1}{2}}}\left(x-1\right) \\ \Rightarrow y-2=-2\left(x-1\right) \\ \Rightarrow y-2=-2x+2 \\ \Rightarrow 2x+y-4=0 \end{gathered}$$
Regards