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.
Solution
The equation of the given curve is x2 = 4y.
Differentiating w.r.t. x, we get
Let (h, k) be the co-ordinates of the point of contact of the normal to the curve x2 = 4y.
Now, slope of the tangent at (h, k) is given by
Hence, slope of the normal at (h, k) = 
Therefore, the equation of normal at (h, k) is 
Since, it passes through the point (1, 2) we have
Now, (h, k) lies on the curve x2 = 4y, so, we have:
...(3)
Solving (2) and (3), we get,
h = 2 and k = 1.
From (1), the required equation of the normal is:
Also, slope of the tangent = 1
Equation of tangent at (1, 2) is:
