Question

b) Find the equation of the rectangula hyperbola whose focus the point (-1, -3) and directrix is the line 2x+y+1=0​

Answer 1

Answer:

Step-by-step explanation:

Given,

$\sf{Focus\,\,\,of\,\,\,the\,\,\,parabola\,,\,\,S\equiv(-1,-3)}$

$\sf{Directrix\,\,\,of\,\,\,the\,\,\,parabola\,,\,\,D:\,2x+y+1=0}$

Let M be the foot of perpendicular on directrix, and P be any point on the parabola

So, for a parabola,

$\tt{SP=PM}$

$\tt{\implies\,\sqrt{(h+1)^2+(k+3)^2}=\dfrac{2h+k+1}{\sqrt{4+1}}}$

$\tt{\implies\,(h+1)^2+(k+3)^2=\dfrac{(2h+k+1)^2}{5}}$

$\tt{\implies\,h^2+2h+1+k^2+6k+9=\dfrac{4h^2+k^2+1+4hk+2k+4h}{5}}$

$\tt{\implies\,h^2+k^2+2h+6k+10=\dfrac{4h^2+k^2+4hk+4h+2k+1}{5}}$

$\tt{\implies\,5h^2+5k^2+10h+30k+50=4h^2+k^2+4hk+4h+2k+1}$

$\tt{\implies\,5h^2-4h^2+5k^2-k^2-4hk+10h-4h+30k-2k+50-1=0}$

$\tt{\implies\,h^2+4k^2-4hk+6h+28k+49=0}$

The required equation of the parabola is

$\boxed{\tt{x^2+4y^2-4xy+6x+28y+49=0}}$