logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the equation of the hyperbola with focus $(2,2) \; e=2$ and directrise $x+y=9$

$\begin{array}{1 1}(A)\;x^2+y^2+4xy-32x-32y+154=0 \\(B)\;x^2-y^2-4xy-32x-32y+154=0 \\(C)\;x^2-y^2-4xy+32x+32y+54=0 \\(D)\;x^2+y^2+4xy+32x+32y+80=0 \end{array}$

Download clay6 mobile app

1 Answer

Let $P(x,y)$ be any point on the hyperbola
$\therefore \large\frac{\text{Distance of P from focus}}{\text{Distance from directrix}}$$=e$
=> $(x-2)^2+(y-2)^2=2^2 \bigg[ \large\frac{x+y-9}{\sqrt 2}\bigg]^2$
$x^2+4-4xy+y^2+4 -4y=4 \bigg[ \large\frac {x+y-81}{2}\bigg]$
$x^2+y^2-4x-4y+8= 2 [x^2+y^2+81+2xy-18y-18x]$
$x^2+y^2-4x-4y+8- 2x^2-2y^2-162-4xy+36y+36x=0$
$-x^2-y^2-4xy+32y+32x-154=0$
$x^2+y^2+4xy-32x-32y+154=0$
Hence A is the correct answer.
answered Apr 10, 2014 by meena.p
 

Related questions

...