Q6 / 40 · NTA Abhyas (JEE Main) 2020 · Mathematics · Hyperbola

If e1 and e2 are the eccentricities of the ellipse x218+y24=1 and the hyperbola x29-y24=1 respectively and e1,e2 is a point on the ellipse 15x2+3y2=k , then the value of k is equal to

  • Correct answer
    A
    16
  • B
    17
  • C
    15
  • D
    14

Solution

e1=1-418=79=73
e1=1+49=139=133

Also, 
15e12+3e22=k⇒k=1579+3139
∴k=16

Hence, the correct answer is option: 'A'.

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