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Find the equation of the hyperbola if focus : $(2 , 3 );$ corresponding directrix : $x+2y=5,e=2$

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  • General equation of a conic with focus at $F(x,y)$,directrix $lx+my+n=0$ and eccentricity $e$ is $(x-x_1)^2+(y-y_1)^2=e^2\bigg[\pm\large\frac{lx+my+n}{\sqrt{l^2+m^2}}\bigg]^2$
  • Which reduces to the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
Step 1:
$F(2,3)$ and directrix $l : x+2y=5,e=2$.
Let $P$ be a point on the hyperbola and $PM\perp l$
$\large\frac{SP^2}{PM^2}=$$e^2$
$SP^2=e^2PM^2$
Step 2:
$(x-2)^2+(y-3)^2=4\bigg(\pm\large\frac{x+2y-5}{\sqrt{1+4}}\bigg)^2$
$5[(x-2)^2+(y-3)^2]=4(x+2y-5)^2$
$5x^2-20x+20+5y^2-30y+45=4x^2+16y^2+100+16xy-40x-80y$
The equation of the hyperbola is $x^2-16xy-11y^2+20x+50y-35=0$
answered Jun 18, 2013 by sreemathi.v
 

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