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  • A line segment of length  moves in such a way that its ends

A line segment of length \mathrm a + \mathrm b moves in such a way that its ends are always on two fixed perpendicular straight lines. Then the locus of the point on this line which divides it into portions of lengths \mathrm a and \mathrm b is

Option: 1

a parabola
 


Option: 2

a circle


Option: 3

an ellipse

 


Option: 4

none of these


Answers (1)

best_answer

\begin{aligned} & \mathrm{B}=((\mathrm{a}+\mathrm{b}) \cos \theta, 0) \mathrm{A} \equiv(0,(\mathrm{a}+\mathrm{b}) \sin \theta) \\ & \mathrm{P}(\mathrm{h}, \mathrm{k}) \equiv\left(\frac{\mathrm{a}(\mathrm{a}+\mathrm{b}) \cos \theta+\mathrm{b} 0}{\mathrm{a}+\mathrm{b}} ; \frac{\mathrm{b}(\mathrm{a}+\mathrm{b}) \sin \theta+0 \cdot \mathrm{a}}{\mathrm{a}+\mathrm{b}}\right) \\ & \text { i.e., } \quad \mathrm{h}=\mathrm{a} \cos \theta ; \mathrm{k}=\mathrm{b} \sin \theta \end{aligned}

\Rightarrow \quad \frac{\mathrm{h}^2}{\mathrm{a}^2}+\frac{\mathrm{k}^2}{\mathrm{~b}^2}=1, so that locus of (\mathrm{h}, \mathrm{k}) is an ellipse.

Hence (c) is the correct answer.

Posted by

Deependra Verma

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