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

Equation to the locus of the point which moves such that the sum of its distances from (−4,3) and (4,3) is 12 is

A
x236+(y3)220=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
x220+(y3)236=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(x3)236+y220=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(x1)236+(y3)220=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B x236+(y3)220=1
We know that locus will be ellipse as sum of it's distances from two fixed points is constant.
Then focii of the ellipse are (±4,3)
So, center of ellipse will be (0,3) and constant distance is 12 which is equal to 2a(length of major axis)
Let e be the eccentricity of ellipse
2a and 2b be the lengths of major and minor axis respectively.
We know that distance between the focii is 2ae=8, then
e=23 and also we know
a2(1e2)=b2b2=20
Now equation of ellipse be x236+(y3)220=1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon