PrepRung
Sign inSign up free

NDA 2023 Maths · Mathematics · Conic Sections · hard

The ellipse with foci (±2,0) and eccentricity 1/2 has the equation

Single correct option

The ellipse with foci (±2,0)(\pm2,0) and eccentricity 12\dfrac12 has the equation
  1. A. x216+y212=1\dfrac{x^2}{16}+\dfrac{y^2}{12}=1
  2. B. x212+y216=1\dfrac{x^2}{12}+\dfrac{y^2}{16}=1
  3. C. x216+y24=1\dfrac{x^2}{16}+\dfrac{y^2}{4}=1
  4. D. x24+y23=1\dfrac{x^2}{4}+\dfrac{y^2}{3}=1
Show answer and solution

Answer: (A) x216+y212=1\dfrac{x^2}{16}+\dfrac{y^2}{12}=1

The foci are at (±ae,0)(\pm ae,0) so ae=2ae=2, and with e=12e=\tfrac12 we get a=4a=4. Then b2=a2(1−e2)=16⋅34=12b^2=a^2(1-e^2)=16\cdot\tfrac34=12. The equation is x216+y212=1\dfrac{x^2}{16}+\dfrac{y^2}{12}=1. Swapping the denominators would put the foci on the yy-axis.
Report a mistake

Original question written in the NDA 2023 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake