Question

Find the equation of normal through the point $(1,2)$ on the parabola $x^2=4y$

Answer 1

$\begin{array}{c}{x}^2=4y\to \left(i\right)\\ differentiate\\ 2x_1=4\frac{d{y}_1}{d{x}_1}\\ \frac{d{y}_1}{d{x}_1}=\frac{x_1}{2}\\ slopeofnormal=-\frac{2}{x_1}\\ normaly-y_1=m\left(x-x_1\right)\\ 2-y_1=\left(-\frac{2}{x_1}\right)\times \left(1-x_1\right)\\ from\left(i\right)\\ \left(-\frac{x_1^2}{4}\right)=\left(2-\frac{2}{x_1}\right)x_1^3\\ x_1=2,y_1=\frac{22}{4}=1\\ normaly-1=\left(-\frac{2}{2}\right)\left(x-2\right)\\ y-1=-x+2\\ x+y=3\end{array}$