Tangents to the curve and parallel to the $y$-axis?

Question

The equation of a curve is $$3x^2 + 8xy + y^2 = -13.$$ Find the equations of the two tangents which are parallel to the $y$-axis.

Answer 1

Hints :

• differentiate (you should get a $f(x,y)dx+g(x,y)dy=0$ equation)
• search the zeros of $\frac {dx}{dy}$ (you should get $y$ in function of $x$)
• replace $y$ (for example) in the initial equation and solve

Answer 2

The equation of any straight line parallel to the y-axis is x=c (where c is some constant).

If this line is tangent to the given curve, the point (c,y) must lie on both the line as well as on the curve, so, $3c^2+8cy+y^2=-13$

$=>y^2+y(8c)+(3c^2+13)=0$, but this is a quadratic equation in y.

As the line is tangent to the given curve, both the value of y should be same($\frac{-8c}{2}=-4c$) to make the two point of intersection coincide.

=>$(8c)^2=4(3c^2+13)$ as the discriminant needs to 0.

$=>c=±1$, so the equation of the required tangents are x=±1.

The point of contact being (c , -4c) ≡ (±1 , ∓4).