wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Consider the parabola x2+4y=0. Let P(a,b) be any fixed point inside the parabola and let S be the focus of parabola. Then the minimum value of SQ+PQ as point Q moves on parabola is:

A
|1a|
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
|ab|+1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
a2+b2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1b
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 1b
Let the foot of perpendicular from Q to the directrix be N.
Now, since focal distance of Q=QN
So, for SQ+PQ to be minimum QN+PQ should also be minimum since SQ=QN
Here, S,P are fixed points and Q,N are variable points.
For QN+PQ to be minimum, P,Q,N should be collinear. So minimum value of SQ+PQ=PQ+QN=PN=1b

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon