Test Index

KEAM 2011 Math Paper

© examsnet.com

Question : 6 of 120

Marks: +1, -0

The line L has intercepts a and b on the coordinate axes. Keeping the origin fixed, the coordinate axes are rotated through a fixed angle. If the line L has intercepts p and q on the rotated axes, then

\frac{1}{a^2}+\frac{1}{b^2}

is equal to

$p^2+q^2$
$p^2-q^2$
$\frac{1}{p^2}+\frac{1}{q^2}$
$\frac{1}{p^2}-\frac{1}{q^2}$
$\frac{1}{q^2}-\frac{1}{p^2}$

Solution:

Since the origin remains same, so the length of the perpendicular from the origin on the line in this positions

\frac{x}{a}+\frac{y}{b}=1

...(1)

and

\frac{x}{p}+\frac{y}{q}=1

....(2)

are equal.

Therefore,

R_{1}=P_{2}

\frac{\left|\frac{0}{a}+\frac{0}{b}-1\right|}{\sqrt{\frac{a}{a^2}+\frac{1}{b^2}}}=\frac{\left|\frac{0}{p}+\frac{0}{q}-1\right|}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}

\Rightarrow \sqrt{\frac{1}{p^2}+\frac{1}{q^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}

\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}

© examsnet.com

Go to Question:

Prev Question

Next Question