b2=a2(1−e2)=a2(1−52)=a253=53a2a2x2+ b2y2=1⇒a29+3a25=1a2=332 b2=532 ∴ Required equation of ellipse 3x2+5y2−32=0
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (-3,1) and has eccentricity √ 2 5 is
JEE Main 2011 — Mathematics Coordinate Geometry
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (−3,1) and has eccentricity 52 is
Held on 30 Apr 2011 · Verified 6 Jul 2026.
Options
- A
5x2+3y2−48=0
- B
3x2+5y2−15=0
- C
5x2+3y2−32=0
- D
3x2+5y2−32=0
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