Conic Sections Question 198
Question: The line $ x-1=0 $ is the directrix of the parabola $ y^{2}-kx+8=0 $ . Then one of the values of k is
[IIT Screening 2000]
Options:
A) $ \frac{1}{8} $
B) 8
C) 4
D) $ \frac{1}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
The parabola is $ y^{2}=4.\frac{k}{4}( x-\frac{8}{k} ) $ . Putting $ y=Y,x-\frac{8}{k}=X, $ the equation is $ Y^{2}=4.\frac{k}{4}.X. $
The directrix is $ X+\frac{k}{4}=0, $ i.e. $ x-\frac{8}{k}+\frac{k}{4}=0 $ . But $ x-1=0 $ is the directrix. So, $ \frac{8}{k}-\frac{k}{4}=1 $
$ \Rightarrow k=-8,4. $
BETA
