Tangents to graphs of implicit relations

Question

Consider the curve given by the equation $xy^2-x^3y=8$. It can be shown that $\dfrac{dy}{dx}=\dfrac{3x^2y-y^2}{2xy-x^3}$

Write the equation of the vertical line that is tangent to the curve.

This is on Khan Academy. I figured out how to solve it based on the solution given. But I have a question. The solution says to use a system of equations:

$$xy^2-x^3y=8\\2xy-x^3=0\\3x^2y-y^2\neq0$$

Why is the last equation, numerator, in the system of equations not equal to zero and the second one, denominator, is equal to zero?

Answer 1

You certainly know that a vertical line $$x=a$$ (either it is a tangent line or not) has undefined slope! So, the numerator must be non-zero but the denominator should be zero and that 's why you are suggested to do a simultaneously system as above.

Answer 2

$$\tan\theta=\dfrac{dy}{dx}=\dfrac{3x^2y-y^2}{2xy-x^3}$$

You want a vertical line to $x-$axis thus $\theta=90$

Thus your denominator must be 0

and your numerator must be non-zero $$xy^2-x^3y=8\\2xy-x^3=0\\3x^2y-y^2\neq0$$

Answer 3

You want the slope of tangent to be infinity so the tangent line is vertical. The fraction $$\dfrac{dy}{dx}=\dfrac{3x^2y-y^2}{2xy-x^3} $$ becomes infinity only if the bottom is zero and the top is not zero.