Find the equation of the normal at a point on the curve x2 = 4y

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409.

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.


The equation of the given curve is x2 = 4y.
Differentiating w.r.t. x, we get
dy over dx equals x over 2
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
right enclose dy over dx end enclose subscript left parenthesis straight h comma space straight k right parenthesis end subscript space equals space straight h over 2
Hence, slope of the normal at (h, k)  = fraction numerator negative 2 over denominator straight h end fraction
Therefore, the equation of normal at (h, k) is 
straight y minus straight k space equals space fraction numerator negative 2 over denominator straight h end fraction left parenthesis straight x minus straight h right parenthesis space space space space space space... left parenthesis 1 right parenthesis
Since, it passes through the point (1, 2) we have
2 minus straight k equals fraction numerator negative 2 over denominator straight h end fraction left parenthesis 1 minus straight h right parenthesis space or space space straight k space equals space 2 plus 2 over straight h left parenthesis 1 minus straight h right parenthesis space space space... left parenthesis 2 right parenthesis
Now, (h, k) lies on the curve x2 = 4y, so, we have:
straight h squared equals 4 straight k                        ...(3) 
Solving (2) and (3),  we get,
h = 2  and k = 1.
From (1), the required equation of the normal is:
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