CameraIcon
CameraIcon
SearchIcon
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 the parabola. Then the minimum value SQ + PQ as point Q moves on the 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 foot of perpendicular from Q to the directrix be N
SQ + PQ = QN + PQ is minimum it P, Q & N are collinear
So minimum value of SQ + PQ = PN = 1 - b
(PN is the distance of P from directrix. It will be equal to y coordinate minus 'a' if the parabola is x2=4ay . Here, a = -1 and y coordinate = b)


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