What is the equation of the tangent at the vertex of parabola $4y^2+6x=8y+7$?

Question

What is the equation of the tangent at the vertex of this parabola? $$4y^2+6x=8y+7$$

I simplified the equation and got $$4(y-1)^2 =-(6x-11)$$ What do I do further?

Answer 1

The equation of your parabola is $-4(y-1)^2=6(x-\tfrac{11}{6})$, therefore the vertex is $V(\tfrac{11}{6},1)$.

Since the parabola is upside down (opening to the left), the equation of the tangent line at the vertex is $x=\tfrac{11}{6}$.