Zaymiey

📐 Mathematics  ·  Conic Sections  ·  JEE

An ellipse has eccentricity 3/5 and its foci at (±3, 0). Find the equation of the ellipse.

Answer: x 2 /25 + y 2 /16 = 1.

  • A x<sup>2</sup>/25 + y<sup>2</sup>/16 = 1
  • B x<sup>2</sup>/16 + y<sup>2</sup>/25 = 1
  • C x<sup>2</sup>/9 + y<sup>2</sup>/25 = 1
  • D x<sup>2</sup>/25 + y<sup>2</sup>/9 = 1

Correct answer: A. x<sup>2</sup>/25 + y<sup>2</sup>/16 = 1

Explanation: Foci at (±3,0) means c=3. Eccentricity e=c/a=3/5, so a=c/e=3÷(3/5)=5; a²=25. Then b²=a²−c²=25−9=16. Equation: x²/25+y²/16=1. Ans: x²/25+y²/16=1.

The Four Conics by EccentricityCircle (e=0)Ellipse (0<e<1)Parabola (e=1)Hyperbola (e>1)

All four conics form a single family distinguished only by eccentricity: a circle is the most "closed" (e=0), an ellipse is an elongated closed curve, a parabola is the borderline open curve (e=1), and a hyperbola has two separate open branches (e>1).

Concept context

Study circles, parabolas, ellipses, and hyperbolas as curves formed by intersecting a plane with a double cone, with their standard equations and key properties.

Read the full Conic Sections notes →