JEE Main20262 April 2026Morning ShiftMathematicsEllipseActual
Let an ellipse x^2 a^2 + y^2 b^2 = 1 , a < b , pass through the point (4, 3) and have eccentricity 5 3 . Then the length of its latus rectum is :
Options
- A4 5 3
- B2 5
- C7 5 3
- D8 5 3
Correct answer
D. 8 5 3
Step-by-step solution
Given the equation of the ellipse is x^2 a^2 + y^2 b^2 = 1 with a The eccentricity is given by e = 5 3 . Since a Substituting the value of e : a^2 = b^2 (1 - 5 9 ) = 4 9 b^2 The ellipse passes through the point (4, 3) , so substituting x = 4 and y = 3 into the equation of the ellipse gives: 16 a^2 + 9 b^2 = 1 Substituting a^2 = 4 9 b^2 into the above equation: 16 4 9 b^2 + 9 b^2 = 1 36 b^2 + 9 b^2 = 1 45 b^2 = 1 b^2 = 45 Now, finding a^2 : a^2 = 4 9 45 = 20 For an ellipse with a Substituting the values of a^2 and b