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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

If p is the length of the perpendicular from the origin to the line $\large\frac{x}{a}+\frac{y}{6}$$=1$ then $\large\frac{1}{a^2}+\frac{1}{b^2}$=?

$\begin{array}{1 1}(a)\;\large\frac{1}{p^2}&(b)\;p^2\\(c)\;0&(d)\;1\end{array}$

1 Answer

p=length of perpendicular from origin to $\large\frac{x}{a}+\frac{y}{6}$$=1$
$p=\large\frac{\mid 0+0-1\mid}{\sqrt{\Large\frac{1}{a^2}+\frac{1}{b^2}}}$
$\Rightarrow \large\frac{1}{\sqrt{\Large\frac{1}{a^2}+\frac{1}{b^2}}}$
$p=\large\frac{1}{\sqrt{\Large\frac{1}{a^2}+\frac{1}{b^2}}}$
$p^2(\large\frac{1}{a^2}+\frac{1}{b^2})$$=1$
$(\large\frac{1}{a^2}+\frac{1}{b^2})$$=\large\frac{1}{p^2}$
Hence (a) is the correct answer.
answered Feb 4, 2014 by sreemathi.v
 

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