The sum of the slopes of the tangent of the parabola \(y^2\)=4ax drawn from the point (2,3) is

Solution :

The equation of tangent to the parabola \(y^2\) = 4ax is y = mx + \(a\over m\).

Since it is drawn from point (2,3)

Therefore it lies on tangent y = mx + \(a\over m\).

\(\implies\) 3 = 2m + \(a\over m\)

\(\implies\) 3m = 2\(m^2\) + a

\(\implies\)  2\(m^2\) – 3m + a = 0

Now, Sum of slopes is \(3\over 2\).       [  \(\because\) sum of roots = \(-b\over a\) ]

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