Find the equation

It is given that equation of curve is y = x⁴ – 6x³ + 13x² – 10x + 5

On differentiating with respect to x, we get

 = 4x³ - 18x² +26x -10

Now, Slope of tangent will be the value of  at x = 0 i.e. 
m₁ = 4(0)³ - 18(0)² + 26(0) - 10 = -10

Therefore, the slope of the tangent at (0, 5) is -10.

Then, the equation of the tangent is

y – 5 = -10(x – 0)

⇒ y – 5 = 10x

⇒ 10x +y =5

Also, slope of normal at (0,5)

Now, equation of the normal at (0,5)

⇒ 10y - 50 = x

⇒ x -10y +50 = 0