Question

# Find the equation of the ellipse satisfying the given condition e=34, foci on Y-axis, centre at origin and passes through (6,4). Or Find the equation of the hyperbola with vertices at (±5,0) and foci (±7,0)

Open in App
Solution

## Let the equation of ellipse be x2a2+y2b2=1 …(i) If the ellipse passes through (6,4) then (6)2a2+(4)2b2=1⇒36a2+16b2=11 …(ii) We know that, e=√1−a2b2 [∵b>a] ⇒ 34=√1−a2b2 ⇒ 916=1−a2b2 ⇒ a2b2=1−916=716 ⇒ a2=716b21 …(iii) From Eqs. (i) and (iii), we get 36×167b2+16b2=1 ⇒ 167b2(36+7)=1 ⇒1b2=716×43 ⇒ b2=16×437 Put the value of b^2 in Eq. (iii) we get a2=716×16×437=43 On substituting the values of a2 and b2 in Eq. (i), we get x243+7y216×43=1 ∴ 16x2+7y2=688, which is the required equation of ellipse. Or Since, the vertices are on the X-axis with origin at the mid-point. ∴ (±a,0)=(±5,0)and(±c,0)=(±7,0) Then, equation of the hyperbola is of the form x2a2−y2b2=1 Since, the vertices are (±5,0) and foci are (±7,0]) ∴ a=5 and c=7 We know that, in hyperbola c2=a2+b2 ⇒72=52+b2⇒b2=49−25=24 On putting the value of b2 in Eq. (i) we get x225−y224=1⇒24x2−25y2=600

Suggest Corrections
0