What does the equation of the straight lines $7x^2+4xy+4y^2=0$ become when the axes are the bisectors of the angles between them?

Question

What does the equation of the straight lines $7x^2+4xy+4y^2=0$ become when the axes are the bisectors of the angles between them?

My Attempt: $$7x^2+4xy+4y^2=0$$ Comparing the above equation with $ax^2+2hxy+by^2=0$ we get $a=7$, $h=2$ and $b=4$.

If $\alpha$ be the angle between the lines represented by the given equation, we write: $$\tan {\alpha} = \dfrac {2\sqrt {h^2-ab}}{a+b}$$ $$\tan {\alpha} = \dfrac {2\sqrt {4-28}}{7+4}$$ $$\tan {\alpha} = \dfrac {2\sqrt {-24}}{11}$$

Edit after Alberto Saracco's comment.

Answer 1

Given the equation

$$7x^2+4xy+4y^2=0$$

try to complete the square, getting

$$0=7x^2+4xy+4y^2=6x^2+(x+2y)^2$$

This shows your equation is a point. For it to hold true, you must have $x=0$ and $x+2y=0$, which means the only point satisfying it is the origin $(0,0)$...

Answer 2

Let me answer the question assuming the equation is

$$7x^2+4xy\color{red}-4y^2=0$$

Let $$(7x^2+4xy-4y^2)=(\alpha_1x+2y)(\alpha_2x-2y)=\alpha_1\alpha_2x^2+2(\alpha_2-\alpha_1)x-4y^2$$

Then, we have $\alpha_1\alpha_2=7$ and $\alpha_2-\alpha_1=2$

$$\alpha_1(2+\alpha_1)=7$$

$$\alpha_1^2+2\alpha_1-7=0$$

$$\alpha_1=\frac{-2\pm \sqrt{4+28}}{2}=-1\pm 2\sqrt{2}$$ $$\alpha_2=1\pm 2\sqrt{2}$$

Hence we have the factorization that $$((-1+2\sqrt{2})x+2y)((1+2\sqrt{2})x-2y)=0$$

The equations of the two lines are

$y = \frac{1-2\sqrt2}{2}x$ and $y=\frac{1+2\sqrt2}{2}x$.

The corresponding vectors for the lines are $v_1=\begin{bmatrix} 2 \\ 1-2\sqrt2\end{bmatrix}$ and $v_2\begin{bmatrix} 2 \\ 1+2\sqrt{2}\end{bmatrix}$

Let the acute angle between them be $\beta$, then we have $$\cos \beta = \frac{|v_1\cdot v_2|}{\|v_1\|\|v_2\|}$$

$$\sec^2 \beta = \frac{\|v_1\|^2\|v_2\|^2}{|v_1 \cdot v_2|^2} = \frac{[4 + (1-2\sqrt{2})^2][4 + (1+2\sqrt{2})^2]}{9}=\frac{(13-4\sqrt{2})(13+4\sqrt{2})}{9}=\frac{137}9$$

We have $$\tan^2 \beta =\sec^2 \beta-1=\frac{128}{9}=\frac{8^2 \cdot 2}{3^2}$$

$$\tan \beta = \frac{8\sqrt2}{3}$$

$$\frac{2\tan \frac{\beta}2}{1-\tan^2 \frac{\beta}{2}}=\frac{8\sqrt2}{3}$$

Let $\gamma = \frac{4\sqrt2}{3}$, then $\gamma^2 = \frac{32}{9}$ and we have

$$\gamma \tan^2 \frac{\beta}{2}+ \tan \frac{\beta}{2}-\gamma = 0$$

$$\tan \frac{\beta}{2}= \frac{-1 \pm \sqrt{1+4\gamma^2} }{2\gamma}$$

Due to $\beta$ being acute, we have

$$\tan \frac{\beta}{2}= \frac{-1 + \sqrt{1+4\gamma^2} }{2\gamma}$$

$$\tan^2 \frac{\beta}{2}=\frac{2+4\gamma^2-2\sqrt{1+4\gamma^2}}{4\gamma^2}=\frac{1+2\gamma^2-\sqrt{1+4\gamma^2}}{2\gamma^2}$$

Note that the angle bisectors are perpendicular, hence the equation using the new coordinate is

$$y^2=\tan^2 \frac{\beta}{2}x^2=\left(\frac{73-3\sqrt{137}}{64}\right)x^2$$