Question

What is the equation of the normal to the curve y = 3x 2 – 7x + 5 at (0, 5)?​

Answer 1

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

$\textsf{Curve is}$

$\mathsf{y=3x^2-7x+5}$

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

$\textsf{The equation of normal to the curve at (0,5)}$

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

$\mathsf{Consider,}$

$\mathsf{y=3x^2-7x+5}$

$\textsf{Differentiate with respect to 'x'}$

$\mathsf{\dfrac{dy}{dx}=6x-7}$

$\textsf{Slope of tangent at (0,5)}$

$\mathsf{m=\left(\dfrac{dy}{dx}\right)_{(0,5)}}$

$\mathsf{m=6(0)-5}$

$\mathsf{m=-5}$

$\implies\textsf{Slope of normal=}\mathsf{\dfrac{-1}{m}=\dfrac{1}{5}}$

$\textsf{Equation of normal is}$

$\mathsf{y-y_1=\dfrac{-1}{m}(x-x_1)}$

$\mathsf{y-5=\dfrac{1}{5}(x-0)}$

$\mathsf{y-5=\dfrac{x}{5}}$

$\mathsf{5y-25=x}$

$\implies\boxed{\mathsf{x-5y+25=0}}$

$\therefore\textbf{The equation of normal is x-5y+25=0}$

Answer 2

x - 7y + 35 = 0 is the  equation of the normal to the curve y = 3x² – 7x + 5 at (0, 5)

Given :

• Curve Equation
• y = 3x² - 7x + 5

To Find:

• Equation of Normal at (0 , 5)

Solution:

Step 1:

Find Derivative wrt x to find slope of tangent at any point on curve

y = 3x² - 7x + 5

dy/dx = 6x - 7

Step 2:

Substitute x=0  to find slope of tangent at (0,5) and solve

dy/dx = 6(0) - 7

dy/dx = -7

Step 3:

Normal is perpendicular to tangent and Product of slopes of two perpendicular lines is - 1

Hence slope of normal = - 1/(-7) = 1/7

Step 4:

Use Equation of of line (y - y₁) = m(x-x₁)

and substitute x₁ = 0 and y₁ = 5  and m = 1/7 and simplify

y - 5  = (1/7)(x - 0)

=> 7y - 35 = x

=> x - 7y + 35 = 0

x - 7y + 35 = 0 is the  equation of the normal to the curve y = 3x² – 7x + 5 at (0, 5)