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Offline Centres

If ‘p’ is the length of the perpendicular drawn from the origin upon a straight line then the locus of mid point of the portion of the line intercepted between the coordinate axes is

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$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{p^{2}}$

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{2}{p^{2}}$

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{4}{p^{2}}$

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{p}$

answer is C.

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Detailed Solution

Equate the distance from $(0, 0)$ to the line $\frac{x}{x_{1}} + \frac{y}{y_{1}} = 2$

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