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Joint Entrance Examination
Mathematics
Ellipse
Question 160
Question 160 of 183
Medium
MCQ
If the curve x2 + 2y2 = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is :
A
\[{\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 4}} \right)\]
B
\[{\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 3}} \right)\]
C
\[{\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 3}} \right)\]
D
\[{\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)\]
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Official solution
Ellipse : \[{x \over 2} + {y \over 1} = 1\] Line :
x
+
y
=
1
x + y = 1
x
+
y
=
1
Using homogenisation
x
2
+
2
y
2
=
2
(
1
)
2
{x^2} + 2{y^2} = 2{(1)^2}
x
2
+
2
y
2
=
2
(
1
)
2
x
2
+
2
y
2
=
2
(
x
+
y
)
2
{x^2} + 2{y^2} = 2{(x + y)^2}
x
2
+
2
y
2
=
2
(
x
+
y
)
2
x
2
+
2
y
2
=
2
x
2
+
2
y
2
+
4
x
y
{x^2} + 2{y^2} = 2{x^2} + 2{y^2} + 4xy
x
2
+
2
y
2
=
2
x
2
+
2
y
2
+
4
x
y
x
2
+
4
x
y
=
0
{x^2} + 4xy = 0
x
2
+
4
x
y
=
0
for
a
x
2
+
2
h
x
y
+
b
y
2
=
0
a{x^2} + 2hxy + b{y^2} = 0
a
x
2
+
2
h
x
y
+
b
y
2
=
0
\[\tan \theta = \left| {{{2\sqrt {{h^2} - ab} } \over {a + b}}} \right|\] \[\tan \theta = \left| {{{2\sqrt {{{(2)}^2} - 0} } \over {1 + 0}}} \right|\] \[\tan \theta = - 4\] \[\cot \theta = - {1 \over 4}\] \[\theta = {\cot ^{ - 1}}\left( { - {1 \over 4}} \right)\] \[\theta = \pi - {\cot ^{ - 1}}\left( {{1 \over 4}} \right)\] \[\theta = \pi - \left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {{1 \over 4}} \right)} \right)\] \[\theta = {\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)\]
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