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Joint Entrance Examination
Mathematics
Hyperbola
Question 61
Question 61 of 106
Medium
MCQ
If
e
1
e_1
e
1
and
e
2
e_2
e
2
are the eccentricities of the ellipse
x
2
18
+
y
2
4
=
1
\dfrac{x^2}{18}+\dfrac{y^2}4=1
18
x
2
+
4
y
2
=
1
and the hyperbola
x
2
9
−
y
2
4
=
1
\dfrac{x^2}9-\dfrac{y^2}4=1
9
x
2
−
4
y
2
=
1
respectively, and
(
e
1
,
e
2
)
(e_1,e_2)
(
e
1
,
e
2
)
is a point on the ellipse
15
x
2
+
3
y
2
=
k
15x^2+3y^2=k
15
x
2
+
3
y
2
=
k
, then
k
k
k
is equal to
A
16
16
16
B
17
17
17
C
15
15
15
D
14
14
14
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Official solution
\[e_1^2=1-\frac4{18}=\frac79,\qquad e_2^2=1+\frac49=\frac{13}9.\] \[k=15e_1^2+3e_2^2=15\cdot\frac79+3\cdot\frac{13}9=\frac{105+39}9=\frac{144}9=16.\]
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