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.

#### Solution

The equation of the given curve is *x*^{2} = 4*y*.

Differentiating *x*^{2} = 4*y* with respect to *x*, we get

`dy/dx=x/2`

Let (*h*, *k*) be the coordinates of the point of contact of the normal to the curve *x*^{2} = 4*y*. Now, slope of the tangent at (*h*, *k*) is given by

`[dy/dx]_(h,k)=h/2`

Hence, slope of the normal at (h,k)=-2/h

Therefore, the equation of normal at (*h*, *k*) is

`y-k=-2/h(x-h).....(1)`

Since it passes through the point (1, 2) we have

`2-k=-2/h(1-h) or k=2+2/h(1-h) ...............(2)`

Since (*h*, *k*) lies on the curve *x*^{2} = 4*y*, we have

`h^2=4k ...............(3)`

From (2) and (3), we have *h* = 2 and *k* = 1. Substituting the values of *h* and *k* in (1), we get the required equation of normal as

`y-1=-2/2(x-2) or x+y=3`

Also, slope of the tangent = 1

∴ Equation of tangent at (1, 2) is *y* − 2 = 1 (*x* − 1)

⇒ *y* = *x* + 1