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

Find the equation of an ellipse whose axes lie along coordinate axes and which passes through (4,3) and (-1,4).

Open in App
Solution

Let the equation of the required ellipse be
x2a2+y2b2=1, where a>b ...(i)
The required ellipse passes through (4,3) and (-1,4)
(4)2a2+(3)2b2=1
16a2+9b2=1
16b2+9a2=a2b2 ...(ii)
and (1)2a2+(4)2b2=1
1a2×16b2=1
b2+16a2=a2b2 ...(iii)
Multiplying equation (iii) by 16, we get
16b2+256a2=16a2b2 ...(iv)
Substracting equation (ii) from equation (iv), we get
256a29a2=16a2b2a2b2
247a2=15a2b2
24715=b2
b2=24715
Putting b2=24715 in equation (iii), we get
24715+16a2=a2×24715
16a2247a215=24715
240a2247a215=24715
7a2=247
a2=2477
Putting a2=2477 and b2=24715 in equation (i), we get,
x22477+y224715=1
7x2247+15y2247=1
This is the equation of the required ellipse.

flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Ellipse and Terminologies
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon