A parabola with latus rectum '4a' slides such that it touches the positive co-ordinate axes. The locus of its focus is

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$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{4}{a^{2}}, x > 0, y > 0$

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{a^{2}}, x > 0, y > 0$

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{2}{a^{2}}, x > 0, y > 0$

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{2 a^{2}}, x > 0, y > 0$

answer is A.

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Detailed Solution

Let the focus be $(h, k)$ then equation of the tangent at vertex is $y = - \frac{k}{h} (x - h) i . e . kx + hy - hk = 0$ its distance from $(h, k) = | \frac{h k}{\sqrt{h^{2} + k^{2}}} | = a ∴$ the locus is $a^{2} (x^{2} + y^{2}) = x^{2} y^{2} i.e. \frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{a^{2}}, x > 0, y > 0$