Equation of normal to the curve at y = $4 x^{2}$ at (a,b) is x+16y-258 = 0 . Find the value of (a+b)

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Solution

We are given the equation of normal. We know that tangent and normal are perpendicular and their slopes hold the relation m1. m2 = -1

Using this we can find the slope of tangent. We know that slope of tangent is nothing but the derivative f’(x1), where x1 is the x coordinate of the point where normal is drawn.

x+16y-258 = 0 is the equation of normal given. Slope of this line is $\frac{- 1}{16}$ .

So, slope of tangent will be 16.

Now, we will find the derivative at the point (a,b) for y = $4 x^{2}$

f’(x) = 8x

$\Rightarrow$ f’(a) = 8a

We got this as 16

$\Rightarrow$ 8a = 16

$\Rightarrow$ a = 2

Substituting a = 2 in the equation of the curve we get $= 4 \times 2^{2} = 16$

So the point (a, b) where the normal is drawn is (2, 16)

$\Rightarrow$ a+b = 18

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