Question

Find the slope of the normal to the curve y=4x^2-14x+5 at x=5.

(a) –\(\frac{1}{26}\)

(b) \(\frac{1}{26}\)

(c) 26

(d) -26

I got this question during an internship interview.

This intriguing question originated from Derivatives Application in section Application of Derivatives of Mathematics – Class 12

Answer 1

Right option is (a) –\(\frac{1}{26}\)

Easy explanation: The slope of the tangent at x=5 is given by:

\(\frac{dy}{dx}\)=8x-14

\(\frac{dy}{dx}\)]x=5=8(5)-14=40-14=26

∴ The slope of the normal to the curve is

\(-\frac{1}{slope\, of \,the \,tangent \,at \,x=5}=-\frac{1}{26}\).