Find the equation of the normal to the curve `x^2=4y`which passes through the point (1, 2).

Explanation

As given,

`x^{2}=4 y`

or `d y / d x=x / 2`


`\therefore` Slope of tangent at `(x_{1}, y_{1})=\frac{x_{1}}{2}`

`\therefore` Slope of Normal `=-\frac{2}{x_{1}}`

Since normal passes through `(1,2)`

`\therefore` Slope of normal line `=\frac{y_{1}-2}{x_{1}-1}=\frac{\frac{x_{1}^{2}}{4}-2}{x_{1}-1}`

or `\quad-\frac{2}{x_{1}}=\frac{x_{1}^{2}-8}{4(x_{1}-1)}`

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