Test Index

BITSAT 2014 Paper

Section: Mathematics

© examsnet.com

Question : 97 of 150

Marks: +1, -0

If

P_1

P_2

be the length of perpendiculars from the origin upon the straight lines x secθ + y cosecθ = a and x cosθ – y sinθ = a

\cos^2 \theta

respectively, then the value of

4P_1^2 + P_2^2

.

$a^2$
$2a^2$
$\frac{a^2}{2}$
$3a^2$

Solution:

We have

P_1

= length of perpendicular from (0, 0) on x secθ + y cosecθ = a

i.e.

P_1

=

\left| \frac{a}{\sqrt{\sec^2 \theta + \csc^2 \theta}} \right|

= |a sinθ cosθ|

=

\left| \frac{a}{2 \sin 2\theta} \right|

or 2

P_1

= |a sin 2 θ|

P_2

= Length of the perpendicular from (0, 0) on

x cos θ - y sin θ = a cos 2θ

P_2

=

\left| \frac{a \cos 2\theta}{\sqrt{\cos^2 \theta + \sin^{\theta}}} \right|

= |a cos 2θ|

Now,

4P_1^2+P-2^2

=

a^2 \sin^2 2\theta + a^2 \cos^2 2\theta

=

a^2

© examsnet.com

Go to Question:

Prev Question

Next Question