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

The normal at a point P on the ellipse x2+4y2=16 meets the x axis at Q. If M is the midpoint of the line segment PQ, then the locus of M intersects the latus rectum of the given ellipse at the points

A
(±352, ±27)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(±352, ±197)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(±23, ±17)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
(±23, ±437)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C (±23, ±17)
Given ellipse is
x216+y24=1 (1)
Let point P is (4cosϕ,2sinϕ)
Normal at point P is given by 4xsecϕ2y cosec ϕ=12


Normal intersects x axis at Q(3cosϕ, 0)

Let mid point of PQ is M(α, β), then
α=3cosϕ+4cosϕ2=72cosϕ
cosϕ=27α
and β=sinϕ

Using cos2ϕ+sin2ϕ=1, we have
449α2+β2=1
449x2+y2=1 (2)

Now, equation of latus rectum of ellipse (1) is
x=±ae
x=±23 (3)

Solving (2) and (3), we get
4849+y2=1
y=±17
Hence points of intersection are (±23,±17)

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