Question

Find the foci of the curve $300x^2+320xy+144y^2-1220x-768y+199=0$.

Answer 1

From the given equation of the curve $300x^2+320xy+144y^2-1220x-768y+199=0$, we have

$A=300,H=160,B=144$

Here, $AB-H^2>0$

Thus, the given curve is of ellipse

To find center of the curve, partially differentiate it w.r.t $x$ and w.r.t $y$

$30x+16y=61$ ........ [Partial differentiation w.r.t $x$]

$10x+9y=24$ ......... [Partial differentiation w.r.t $y$]

Solving for $x$ and $y$, we get

Center of the ellipse as $(1.5,1)$

Simplify equation by knowing the center

Take $x=X+1.5$

and $y=Y+1$

Thus, the equation now becomes

$300X^2+320XY+144Y^2=1100$

We need to eliminate the term $XY$

For eliminating the term $XY,$ we need to find the degree of rotation of the axis

i.e. $\tan 2\theta =\frac{H}{A-B}$

i.e. $\tan 2\theta =\frac{320}{300-144}$

$\Rightarrow \theta ={32}^o$

$X=u\cos \theta -v\sin \theta$

$Y=u\sin \theta +v\cos \theta$

Substituting the value of $\theta$ and expressing the equation in the form of $u$ and $v$.

The equation will be:

$\frac{u^2}{2.76}+\frac{v^2}{25}=1$

Now the foci can be easily find as $(-1,5)$ and$(4,-3)$.