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

Two tangents to the ellipse x2b2+y2a2=1 make angles θ1,θ2 with the major axis. If tanθ1,tanθ2 is a constant k then the equation to the locus of their point of intersection is

A
y2b2=kxy
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
y2+b2=kxy
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
y2b2=k(x2a2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
y2+b2=k(x2+a2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A y2b2=kxy
The condition is not clearly specified in the question. Let us assume the condition as cotθ1+cotθ2=k

Given equation of the ellipse is x2b2+y2a2=1

Equations of the tangents to this ellipse are given by,
y=mx±a2m2+b2

Let coordinates of point of intersection of these tangents are (h,k)
Thus, this point must satisfy equations of both the tangents

Thus, put x=h and y=k in above equations, we get,
k=mh±a2m2+b2
kmh=±a2m2+b2

Squaring both sides, we get,
(kmh)2=a2m2+b2
k22mkh+m2h2=a2m2+b2
a2m2m2h2+2mkh+b2k2=0
(a2h2)+(2kh)m+(b2k2)=0

This is a quadratic equation in m.

Now, given cotθ1+cotθ2=k

1tanθ1+1tanθ2=k

1m1+1m2=k

m1+m2m1m2=k

sum of rootsproduct of roots=k

(2kha2h2)(b2k2a2h2)=k

2khb2k2=k

2kh=k(b2k2)

Replace h by x and k by y, we get,
2yx=k(b2y2)
2yx=k(y2b2)

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