Conic Sections Question 357
Question: $ x^{2}-4y^{2}-2x+16y-40=0 $ represents
[DCE 1999]
Options:
A) A pair of straight lines
B) An ellipse
C) A hyperbola
D) A parabola
Show Answer
Answer:
Correct Answer: C
Solution:
$ x^{2}-2x-4y^{2}+16y-40=0 $
therefore $ (x^{2}-2x)-4(y^{2}-4y)-40=0 $
therefore $ {{(x-1)}^{2}}-1-4[{{(y-2)}^{2}}-4]-40=0 $
therefore $ {{(x-1)}^{2}}-4{{(y-2)}^{2}}=25 $
therefore $ \frac{{{(x-1)}^{2}}}{25}-\frac{{{(y-2)}^{2}}}{25/4}=1 $ , which is a hyperbola.
BETA
