Question

Find the case of tangents & normal to the curve $y=x^4-6x^3+13x^2-10x+5$ at $(1,3)$

• A. $y+x-7=0$
• B. $2y+x-7=0$
• C. $2y+x+7=0$
• D. None of these

Answer 1

The correct option is B $2y+x-7=0$

$y=x^4-6x^3+13x^2-10x+5$

$\therefore \frac{dy}{dx}=4x^3-18x^2+26x-10$

$\therefore \frac{dy}{dx}=4(1{)}^3-18(1{)}^2+26\left(1\right)-10$

$(x=1,y=3)$

$\therefore \frac{dy}{dx}=4-18+26-10=2$

$\therefore$ Equation of Tangent:-

$y-3=2(x-1)$

$\therefore y-3=2x-2$

$\therefore y-2x-1=0$

$\therefore$ Equation of Normal:-

$y-3=\frac{-1}{2}(x-1)$

$\therefore 2y-6=-x+1$

$\therefore 2y+x-7=0$.