Question

Find the foci and the eccentricity of the conics $4xy-3x^2-2ay=0.$

Answer 1

The given equation is
$4xy-3x^2-2xy=0$
Differentiating partially w.r.t. x and y, one by one, we get
$\frac{\partial f}{\partial x}=3x-2y=0$ and $\frac{\partial f}{\partial x}=2x-a=0$
Where,
$x=\frac{a}{2},y=\frac{3a}{4}$ and $c^{{}^{\prime }}=\frac{-3a^2}{4}$
So the equation referred to parallel axes through the center is
$3x^2-4xy=-\frac{-3a^2}{4}$
Now, $\tan \theta =\frac{2n}{a-b}$ or $\frac{2\tan \theta }{1-{\tan }^2\theta }=-\frac{2}{3}$
or, $2{\tan }^2\theta -3\tan \theta -2=0$
Hence either $\tan \theta =2$ or $-\frac{1}{2}$
Say $\tan {\theta }_1$ and $\tan {\theta }_2$
Again, $r^2=\frac{-\frac{3}{4}{a}^2(1+{\tan }^2\theta )}{3-4\tan \theta }=\frac{3}{4}{a}^2$ or $-\frac{3}{16}{a}^2$
As $(\tan \theta =2or-\frac{1}{2})$
So, $r_1^2-r_2^2=\frac{15}{16}{a}^2,\cos {\theta }_1=\frac{1}{\sqrt{5}},\sin {\theta }_1=\frac{2}{\sqrt{5}}$
So the co ordinates of foci are
$(\frac{a}{2}\pm \frac{a}{4}\sqrt{3},\frac{3a}{4}\pm \frac{a}{2}\sqrt{3})$
by$(x\pm \sqrt{r_1^2-r_2^2}\cos {\theta }_1,y\pm \sqrt{r_1^2-r_2^2}\sin {\theta }_1)$
and $e^2=\frac{{\alpha }^2-{\beta }^2}{{\alpha }^2}=\frac{\frac{3}{4}+\frac{3}{16}}{\frac{3}{4}}=\frac{5}{4}$
$e=\frac{1}{2}\sqrt{5}$