Let e1 and e2 are the eccentricities of the ellipse (x2/18)+(y2/4)=1 and the hyperbola (x2/9)-(y2/4)=1 respectively. If (e1 , e2) is a point on the ellipse 15x2+3y2=k , then the value of k is equal to

Question

Let e1 and e2 are the eccentricities of the ellipse (x2/18)+(y2/4)=1 and the hyperbola (x2/9)-(y2/4)=1 respectively. If (e1 , e2) is a point on the ellipse 15x2+3y2=k , then the value of k is equal to (A) 16 (B) 17 (C) 15 (D) 14

Tardigrade Admin · Accepted Answer

e1=√1 - (4/18)=√(7/9)=(√7/3) e1=√1 + (4/9)=√(13/9)=(√13 /3) Also, 15e12+3e22=k⇒ k=15((7/9))+3((13/9)) ∴ k=16

Q. Let $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse $\frac{x ^{2}}{18} + \frac{y ^{2}}{4} = 1$ and the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ respectively. If $(e_{1}, e_{2})$ is a point on the ellipse $15 x^{2} + 3 y^{2} = k$ , then the value of $k$ is equal to

$16$

$17$

$15$

$14$

$e_{1} = 1 - \frac{4}{18} = \frac{7}{9} = \frac{7}{3}$
$e_{1} = 1 + \frac{4}{9} = \frac{13}{9} = \frac{13}{3}$
Also,
$15 e_{1}^{2} + 3 e_{2}^{2} = k \Rightarrow k = 15 (\frac{7}{9}) + 3 (\frac{13}{9})$
$∴ k = 16$

Q. Let e1​ and e2​ are the eccentricities of the ellipse 18x2​+4y2​=1 and the hyperbola 9x2​−4y2​=1 respectively. If (e1​,e2​) is a point on the ellipse 15x2+3y2=k , then the value of k is equal to

16

17

15

14