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Question

The equation to the locus of the middle points of the portion of the tangent to the ellipse $$\displaystyle {\frac{x^2}{16}\, +\, \frac{y^2}{9}\, =\, 1}$$ included between the co-ordinate axes is the curve.


A
9x2+16y2=4x2y2
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B
16x2+9y2=4x2y2
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C
3x2+4y2=4x2y2
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D
9x2+16y2=x2y2
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Solution

The correct option is A $$9x^2\, +\, 16y^2\, =\, 4x^2y^2$$
Let any tangent of ellipse is
$$\displaystyle {\frac{x\, \cos \theta}{4}\, +\, \frac{y\, \sin \theta}{3}\, =\, 1}$$

Let it meets axes at $$\displaystyle {A \left (\frac{4}{\cos \theta},\, 0 \right )\, \&\, B \left (0,\, \frac{3}{\sin \theta} \right )}$$

Let midpoint of $$AB$$ is $$(h, k)$$ then

$$2h\, =\, \displaystyle \frac{4}{\cos \theta},\, 2k\, =\, \displaystyle \frac{3}{\sin \theta}$$

$$\because \cos^2 \theta\, +\, \sin^2 \theta\, =\, 1$$

$$\therefore\, \displaystyle {\frac{16}{4h^2}\, +\, \frac{9}{4k^2}\, =\, 1}$$

$$\Rightarrow\, 16k^2\, +\, 9h^2\, =\, 4h^2k^2$$

Hence, locus is $$16y^2\, +\, 9x^2\, =\, 4x^2y^2$$

Maths

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