Question

Find the equation of the tangent and normal lines to the curve y3-xy+x2=1 at (1,0)

Answer 1

$$y^3-xy+x^2=1$$

Differentiate both sides with respect to $x$

$$3y^2y'-y-xy'+2x=0$$

$$y'=\dfrac{y-2x}{3y^2-x}$$

$$slope=m=\dfrac{0-2(1)}{3(0)^2-1}=2$$

The tangent line in point-slope form is

$$y-0=2(x-1)$$

The tangent line in slope-intercept form is

$$y=2x-2$$

For the normal line

$$slope_2=m_2=-\dfrac{1}{m}=-\dfrac{1}{2}$$

The normal line in point-slope form is

$$y-0=-\dfrac{1}{2}(x-1)$$

The normal line in slope-intercept form is

$$y=-\dfrac{1}{2}x+\dfrac{1}{2}$$