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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 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
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