Conic Sections Question 368
Question: The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse $ x^{2}+2y^{2}=2 $ between the co-ordinates axes, is
[IIT Screening 2004]
Options:
A) $ \frac{1}{x^{2}}+\frac{1}{2y^{2}}=1 $
B) $ \frac{1}{4x^{2}}+\frac{1}{2y^{2}}=1 $
C) $ \frac{1}{2x^{2}}+\frac{1}{4y^{2}}=1 $
D) $ \frac{1}{2x^{2}}+\frac{1}{y^{2}}=1 $
Show Answer
Answer:
Correct Answer: C
Solution:
Let the point of contact be $ R\equiv (\sqrt{2}\cos \theta ,\sin \theta ) $
Equation of tangent AB is $ \frac{x}{\sqrt{2}}\cos \theta +y\sin \theta =1 $
$ \Rightarrow $ $ A\equiv (\sqrt{2}\sec \theta ,0);B\equiv (0,\text{cosec }\theta ) $
Let the middle point Q of AB be $ (h,k) $
$ \Rightarrow $ $ h=\frac{\sec \theta }{\sqrt{2}},k=\frac{\text{cosec }\theta }{2}\Rightarrow \cos \theta =\frac{1}{h\sqrt{2}},\sin \theta =\frac{1}{2k} $
$ \Rightarrow $ $ \frac{1}{2h^{2}}+\frac{1}{4k^{2}}=1 $ , \Required locus is $ \frac{1}{2x^{2}}+\frac{1}{4y^{2}}=1 $ .
Trick : The locus of mid-points of the portion of tangents to the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ intercepted between axes is $ a^{2}y^{2}+b^{2}x^{2}=4x^{2}y^{2} $
i.e., $ \frac{a^{2}}{4x^{2}}+\frac{b^{2}}{4y^{2}}=1 $ or $ \frac{1}{2x^{2}}+\frac{1}{4y^{2}}=1 $ .
BETA
