Question

Find the equation of the normal at a point on the curve x² = 4y which passes through the point (1, 2). Also find the equation of the corresponding tangent.

Answer 1

The equation of the given curve is x² = 4y.

Differentiating x² = 4y 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 x² = 4y. 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 x² = 4y, 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