Question

The slope of the perpendicular to the tangent to curve y = x2 – 5 at x = 1 is​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{Curve\;is\;y=x^2-5}$

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

$\textsf{Slope of perpendicular to the tangent to the given}$

$\textsf{curve at x=1}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,\;\;y=x^2-5}$

$\textsf{Differentiate with respect to x}$

$\mathsf{\dfrac{dy}{dx}=2x}$

$\textbf{Slope of tangent at x=1}$

$\mathsf{=\left(\dfrac{dy}{dx}\right)_{x=1}}$

$\mathsf{=2(1)}$

$\mathsf{=2}$

$\mathsf{Slope\;of\;normal=\dfrac{-1}{Slope\;of\;tangent}}$

$\implies\boxed{\mathsf{Slope\;of\;normal=\dfrac{-1}{2}}}$

$\underline{\textbf{Find more:}}$

Find the angle of tangent drawn

to the curve y=3x^2-7x +5 at the point (1,1)with the x axis​

https://brainly.in/question/22812663  

Answer 2

Answer:

y=x^2-5

dy/dx=2x

x=1

dy/dx=2

slope of normal=-1/slope of tangent

=-1/2