Question

: Find points on the curve given by y = x3 – 6x2 + x + 3 where the tangents are parallel to the line
y = x + 5.​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Curve is}\;\mathsf{y=x^3-6x^2+x+3}$

$\underline{\textsf{To find:}}$

$\textsf{The points on the given curve where the tangents are paralle to}$

$\mathsf{y=x+5}$

$\underline{\textsf{Solution:}}$

$\textsf{Let (a,b) be the point on the curve at which the tangent}$

$\textsf{is parallel to x-y+5=0}$

$\textsf{Consider,}$

$\mathsf{y=x^3-6x^2+x+3}$

$\textsf{Differentiate with respect to x}$

$\mathsf{\dfrac{dy}{dx}=3x^2-12x+1}$

$\textsf{Slope of tangent}$

$\mathsf{=(\dfrac{dy}{dx})_(a,b)}$

$\mathsf{=3a^2-12a+1}$

$\textsf{Since the tangent is parallel to x-y+5=0, we have}$

$\textsf{Slope of tangent at (a,b)=slope of x-y+5=0}$

$\mathsf{3a^2-12a+1=1}$

$\mathsf{3a^2-12a=0}$

$\mathsf{3a(a-4)=0}$

$\implies\mathsf{a=0,4}$

$\textsf{Since (a,b) lies on the curve, we have}\;\mathsf{b=a^3-6a^2+a+3}$

$\textsf{when a=0,}\;\mathsf{b=0^3-6(0)^2+0+3=3}$

$\textsf{when a=4,}\;\mathsf{b=4^3-6(4)^2+4+3=64-96+4+3=-25}$

$\therefore\textsf{The required points on the curve are (0,3) and (4,-25)}$