Equation of normal to the curve at y = 4x2 at (a,b) is x+16y-258 = 0 . Find the value of (a+b)
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 −116.
So, slope of tangent will be 16.
Now, we will find the derivative at the point (a,b) for y = 4x2
f’(x) = 8x
⇒ f’(a) = 8a
We got this as 16
⇒ 8a = 16
⇒ a = 2
Substituting a = 2 in the equation of the curve we get =4×22=16
So the point (a, b) where the normal is drawn is (2, 16)
⇒ a+b = 18