If $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ touches the circle $x^{2} + y^{2} = a^{2}$, the point $(1/\alpha,\ 1/\beta)$ lies on a/an

Question

Dr. Abubacker · Accepted Answer

(b)
Since, the perpendicular distance from centre (0, 0) to be tangent =radius of the circle
$\Rightarrow \ \ \frac{| - 1|}{\sqrt{\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}}} = a$
$\Rightarrow \frac{1}{a^{2}} = \frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}$
The locus of $\left( \frac{1}{\alpha},\frac{1}{\beta} \right)$ is $\frac{1}{a^{2}} = \frac{1}{x^{2}} + \frac{1}{y^{2}}$

Q. If $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ touches the circle $x^{2} + y^{2} = a^{2}$, the point $(1/\alpha,\ 1/\beta)$ lies on a/an

Explanation

Reviewed by Dr. Abubacker

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