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

On the parabola y2=64x, find the point nearest to the straight line 4x+3y−14=0.

A
24,9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
9,12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
9,24
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
9,24
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is C 9,24
Given parabola is y2=64x ......... (i)
The line is 4x+3y14=0 ......... (ii)
Any point on the parabola will have the form (t264,t)
And points on the line will be of the form (143t4,t)
Distance between (t264,t) and (143t4,t) is
D=(143t4t264)2+(tt)2
D=143t4t264
For the nearest distance, D=0
342t64=0
t=24y=24
At y=24,x=9 on the parabola
Thus, the point nearest to the straight line is (9,24).

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Hyperbola and Terminologies
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon