Finding the vertex of the parabola $x^2+4xy+4y^2+2y+2x+75=0$ using differentiation

Question

So my math teacher in school just gave me the equation of a parabola: $$x^2+4xy+4y^2+2y+2x+75=0$$ I am asked to find its vertex using differentiation and not using any vertex form or anything like that. How do I do that?

Answer 1

Particular procedure when given $f = x^2+4xy+4y^2+2y+2x+75 = 0$

$$ f_x = 2x+4y+2 =0\\ f_y = 4x+8y+2 = 0 $$

represent parallel lines. The parabola symmetry axis has the same gradient so defining now

$$ l\to -4x+2y-\lambda $$

which is perpendicular to $f_x, f_y$ the next step is to determine $\lambda$ imposing that $l$ is tangent do $f=0$ so substituting $y = \frac 12(4x-\lambda)$ into $f=0$ and solving for $x$ we get

$$ x = \frac{1}{25}\left(5\lambda-3\pm\sqrt{-1866-5\lambda}\right) $$

At tangency, $x$ is unique so $-1866-5\lambda = 0$ giving $\lambda = -\frac{1866}{5}$ and one of the tangents is

$$ l = -4x+2y-\frac{1866}{5}=0 $$

The vertex determination follows.

$$ x^* = -\frac{1869}{25},\ \ y^* = \frac{927}{25} $$