Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
For the parabola y2 = 4x, the point P whose focal distance is 17, is
Question Error Report
Question is incomplete/wrong
Question not belongs to this Chapter
Answer is wrong
Solution is wrong
Answer & Solution is not matching
Spelling mistake
Image missing
Website not working properly
Other (not listed above)
Error description
Thank you for reporting, we will resolve it shortly
Back to Question
Thank you for reporting, we will resolve it shortly
Q. For the parabola $y^2 = 4x$, the point $P$ whose focal distance is $17$, is
KCET
KCET 2009
Conic Sections
A
(16, 8) or (16,-8)
53%
B
(8, 8) or (8,-8)
19%
C
(4, 8) or (4,-8)
19%
D
(2, 8) or (2,-8)
9%
Solution:
Given, $y^{2}=4 x$
Let $P(h, k)$ be any point on the parabola
$\therefore \,\,\,\,\,\,\,(h-1)^{2}+(k-0)^{2}=17^{2}$
Also, $k^{2}=4 h$
$\therefore \,\,\,\,\,\,\,h^{2}+1-2 h+4 h=289$
$\Rightarrow \,\,\,\,\,\,\, h^{2}+2 h-288=0$
$\Rightarrow \,\,\,\,\,\,\,(h+18)(h-16)=0$
$\Rightarrow \,\,\,\,\,\,\, h=16 \,\,\,\,\,\,\,(\because h$ cannot be negative)
$\therefore \,\,\,\,\,\,\, k^{2}=64$
$\Rightarrow \,\,\,\,\,\,\, k=\pm 8$
$\therefore $ Points are $(16,8)$ or $(16,-8)$