Question

Find the foci of the curve $16x^2-24xy+9y^2+28x+14y+21=0$.

Answer 1

We complete the square of the first $3$ terms to get $(4x-3y{)}^2$

Thus, we take a new co-ordinate system:

$X=4x-3y$ .... (1)

$Y=3x+4y$ .... (2)

Thus, the curve can be written as:

$X^2+7(4x+2y+3)=0$

$\Rightarrow X^2+\frac{14X}{5}+\frac{28X}{5}+21=0$

$\Rightarrow {(X+\frac{7}{5})}^2=\frac{-28}{5}\times (Y+\frac{11}{10})$

Thus Focus at $(X,Y)=(\frac{-7}{5},\frac{-11}{10})$

Now use equation (1) and (2) to get $(x,y)$

On solving, we get

$(x,y)=(\frac{-4}{5},\frac{-3}{5})$