JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 18)
If e1 and e2 are the eccentricities of the ellipse,
$${{{x^2}} \over {18}} + {{{y^2}} \over 4} = 1$$ and the hyperbola, $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ respectively and (e1, e2) is a point on the ellipse,
15x2 + 3y2 = k, then k is equal to :
17
16
15
14
Explanation
e1 = $$\sqrt {1 - {4 \over {18}}} $$ = $${{\sqrt 7 } \over 3}$$
e1 = $$\sqrt {1 + {4 \over 9}} $$ = $${{\sqrt {13} } \over 3}$$
$$ \because $$ (e1, e2 ) lies on 15x2 + 3y2 = k
$$ \therefore $$ $$15\left( {{7 \over 9}} \right) + 3\left( {{{13} \over 9}} \right)$$ = k
$$ \Rightarrow $$ k = 16
e1 = $$\sqrt {1 + {4 \over 9}} $$ = $${{\sqrt {13} } \over 3}$$
$$ \because $$ (e1, e2 ) lies on 15x2 + 3y2 = k
$$ \therefore $$ $$15\left( {{7 \over 9}} \right) + 3\left( {{{13} \over 9}} \right)$$ = k
$$ \Rightarrow $$ k = 16
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