Conic Sections Question 72
Question: The equation of the hyperbola whose directrix is $ x+2y=1 $ , focus (2, 1) and eccentricity 2 will be
[MP PET 1988, 89]
Options:
A) $ x^{2}-16xy-11y^{2}-12x+6y+21=0 $
B) $ 3x^{2}+16xy+15y^{2}-4x-14y-1=0 $
C) $ x^{2}+16xy+11y^{2}-12x-6y+21=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{(x-2)}^{2}}+{{(y-1)}^{2}}=4[ \frac{{{(x+2y-1)}^{2}}}{5} ] $
therefore $ 5[x^{2}+y^{2}-4x-2y+5] $
$ =4[x^{2}+4y^{2}+1+4xy-2x-4y] $
therefore $ x^{2}-11y^{2}-16xy-12x+6y+21=0 $ .
BETA
