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

A pair of tangents are drawn to the parabola y2=4ax which are equally inclined to a straight line y=mx+c, whose inclination to the axis is α then locus of their point of intersection is

A
y=(xa)tan2α
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
y=(x+a)tan2α
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
y=(xa)tan2α
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
y=(x+a)tanα
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A y=(xa)tan2α
Let the angle between the line and the tangents is θ and α1 & α2 are the incilnation of the tangents to the axis

αα1=θ=α2α2α=α1+α2tan2α=m1+m21m1m2
Let (h,k) be the point from which pair of tangents are drawn
(h,k) lies on the
y=mx+amm2hmk+a=0m1+m2=kh, m1m2=ahtan2α=kh1ahy=(xa)tan2α

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