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

The intersection angle of between the ellipse x2a2+y2b2=1 and the circle x2+y2=ab is:

A
tan1(b+a)ab
No worries! Weā€˜ve got your back. Try BYJUā€˜S free classes today!
B
tan1(abab)
No worries! Weā€˜ve got your back. Try BYJUā€˜S free classes today!
C
tan1(a+bab)
No worries! Weā€˜ve got your back. Try BYJUā€˜S free classes today!
D
tan1(ba)ab
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D tan1(ba)ab
Let us find the point of intersection of ellipse x2a2+y2b2=1 ...(1)
and circle x2+y2=ab ...(2)
We have, x2a2+1b2(abx2)=1x2(1a21b2)=1ab=bab
x2(b2a2)a2b2=babx2=a2ba+b ...(3)
and y2=abx2y2=aba2ba+b (Using (1) and (3))
y2=ab(a+ba)a+b=ab2a+b
Intersection point is a2ba+b,ab2a+b
The slope of tangent to ellipse at any point (x1,y1) is
m1=b2x1a2y1=b2a2a2ba+bab2a+b=b2a2ab
and slope of tangent to the circle is
m2=x1y1=ab
θ=tan1⎜ ⎜ab+b2a2ab1+b2a2.ab⎟ ⎟=tan1(ba)ab

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