If frac{x}{alpha } + frac{y}{beta } = 1 touch | vedclass

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Question

(b) $y = - \frac{\beta }{\alpha }x + \beta $ touches the circle,

 ${\beta ^2} = {a^2}\left( {1 + \frac{{{\beta ^2}}}{{{\alpha ^2}}}} \right)$

Vedclass · Accepted Answer

Circle

Vedclass · Answer

(b) $y = - \frac{\beta }{\alpha }x + \beta $ touches the circle,

 ${\beta ^2} = {a^2}\left( {1 + \frac{{{\beta ^2}}}{{{\alpha ^2}}}} \right)$

==> $\frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}}$

$= \frac{1}{{{a^2}}}$

 Locus of $\left( {\frac{1}{\alpha },\frac{1}{\beta }} \right)$ is ${x^2} + {y^2}$

$= {\left( {\frac{1}{a}} \right)^2}$.

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

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