Find the equation of the tangent to the curve which is parallel to the line 4x-2y+5=0. Also, write the equation of normal to the curve at the point of contact.

Given that , we have to find tangent to curve which is parallel to the line 4x-2y+5=0.

Differentiating y with respect to x, we get

Now at (x₁,y₁) the equation will be:



The slope of the tangent at (x₁,y₁ ) is:



Now,

The slope of the tangent = slope of the line

⇒ 6(4y - 3) = 48x - 41

⇒ 24y - 18 = 48x - 41

⇒ 48x - 24y - 23 = 0

Equation of normal is:











96y - 72 = 41 - 48x

⇒48x + 96y - 113 = 0