Question

# Tangent drawn to the ellipse $$\dfrac {x^{2}}{a^{2}}+\dfrac {y^{2}}{b^{2}}=1$$ at point $$'P'$$ meets the coordinate axes at point $$A$$ and $$B$$ respectively. Locus of mid-night of segment $$AB$$ is

A
x2a2+y2b2=2
B
a2x2+b2y2=2
C
a2x2+b2y2=4
D
x2a2+y2b2=4

Solution

## The correct option is A $$\dfrac {x^{2}}{a^{2}}+\dfrac {y^{2}}{b^{2}}=2$$$$\frac { { { x^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { { y^{ 2 } } } }{ { { b^{ 2 } } } } =1$$let the point P $$\left( { a\cos \theta ,b\sin \theta } \right)$$$$\begin{array}{l} \frac { x }{ a } \cos \theta +\frac { y }{ b } \sin \theta =1 \\ A=\left( { \frac { a }{ { \cos \theta } } ,0 } \right) \, \, \& \, \, B\left( { 0,\frac { b }{ { \sin \theta } } } \right) \\ let\, \, the\, \, po{ { int } }\, \, \left( { h,k } \right) =\left( { \frac { a }{ { 2\cos \theta } } ,\frac { b }{ { 2\sin \theta } } } \right) \\ comparing\, \, then \\ h=\frac { a }{ { 2\cos \theta } } ,k=\frac { b }{ { 2\sin \theta } } \\ \cos \theta =\frac { a }{ { 2h } } \, \, \& \, \, k=\frac { b }{ { 2k } } \\ now, \\ \frac { { { a^{ 2 } } } }{ { 4{ x^{ 2 } } } } ++\frac { { { b^{ 2 } } } }{ { h{ y^{ 2 } } } } =1 \\ locus=\frac { { { a^{ 2 } } } }{ { { x^{ 2 } } } } +\frac { { { b^{ 2 } } } }{ { { y^{ 2 } } } } =4 \end{array}$$Maths

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