Conic Sections Question 25
Question: The focus of the parabola $ 4y^{2}-6x-4y=5 $ is
[RPET 1997]
Options:
A) (-8/5, 2)
B) (-5/8, 1/2)
C) (1/2, 5/8)
D) (5/8, -1/2)
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation of parabola written in standard form, we get $ 4{{( y-\frac{1}{2} )}^{2}}=6(x+1)\Rightarrow {{( y-\frac{1}{2} )}^{2}}=\frac{3}{2}(x+1)\Rightarrow Y^{2}=\frac{3}{2}X $
where, $ Y=y-\frac{1}{2},X=x+1 $
$ \therefore y=Y+\frac{1}{2},x=X-1 $ ……(i)
For focus $ X=a,Y=0 $
$ \because 4a=\frac{3}{2}\Rightarrow a=\frac{3}{8}\Rightarrow x=\frac{3}{8}-1=-\frac{5}{8} $
$ y=0+\frac{1}{2}=\frac{1}{2} $ , Focus $ =( -\frac{5}{8},\frac{1}{2} ) $ .
BETA
