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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) -26I got this question during an internship interview.This intriguing question originated from Derivatives Application in section Application of Derivatives of Mathematics – Class 12

Answer»

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

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



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