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Joint Entrance Examination
Mathematics
Ellipse
Question 169
Question 169 of 183
Medium
MCQ
The acute angle between the pair of tangents drawn to the ellipse
2
x
2
+
3
y
2
=
5
2 x^{2}+3 y^{2}=5
2
x
2
+
3
y
2
=
5
from the point
(
1
,
3
)
(1,3)
(
1
,
3
)
is :
A
\[\tan ^{-1}\left(\frac{16}{7 \sqrt{5}}\right)\]
B
\[\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)\]
C
\[\tan ^{-1}\left(\frac{32}{7 \sqrt{5}}\right)\]
D
\[\tan ^{-1}\left(\frac{3+8 \sqrt{5}}{35}\right)\]
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Official solution
2
x
2
+
3
y
2
=
5
2{x^2} + 3{y^2} = 5
2
x
2
+
3
y
2
=
5
Equation of tangent having slope m. \[y = mx\, \pm \,\sqrt {{5 \over 2}{m^2} + {5 \over 3}} \] which passes through
(
1
,
3
)
(1,3)
(
1
,
3
)
\[3 = m\, \pm \sqrt {{5 \over 2}{m^2} + {5 \over 3}} \] \[{5 \over 2}{m^2} + {5 \over 3} = 9 + {m^2} - 6m\] \[{3 \over 2}{m^2} + 6m - {{22} \over 3} = 0\]
9
m
2
+
36
m
−
44
=
0
9{m^2} + 36m - 44 = 0
9
m
2
+
36
m
−
44
=
0
\[{m_1} + {m_2} = - 4,\,{m_1}{m_2} = - {{44} \over 9}\] \[{({m_1} - {m_2})^2} = 16 + 4 \times {{44} \over 9} = {{320} \over 9}\] Acute angle between the tangents is given by \[\alpha = {\tan ^{ - 1}}\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|\] \[ = {\tan ^{ - 1}}\left| {{{{{8\sqrt 5 } \over 3}} \over {1 - {{44} \over 9}}}} \right|\] \[ = {\tan ^{ - 1}}\left( {{{24\sqrt 5 } \over {35}}} \right)\] \[\alpha = {\tan ^{ - 1}}\left( {{{24} \over {7\sqrt 5 }}} \right)\]
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