If e 1 and e 2 are the eccentricities of the ellipse x 2 18 + y 2 4 = 1 and the hyperbola x 2 9 - y 2 4 = 1…

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
  1. 16
  2. 17
  3. 15
  4. 14

Solution

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

Also, 
15e12+3e22=kk=1579+3139
k=16

Asked in: JEE Main 2020 (09 Jan Shift 1)

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