Question

If $p$ and $q$ are the perpendicular distances from origin to the straight lines $x\sec \theta -ycosec\theta =a$ and $x\cos \theta +y\sin \theta =a\cos 2\theta$,

then which of the following is correct?

• A. $4p^2+q^2=a^2$
• B. $p^2+q^2=a^2$
• C. $p^2+2q^2=a^2$
• D. $4p^2+q^2=2a^2$

Answer 1

The correct option is A $4p^2+q^2=a^2$

$x\sec \theta -ycosec\theta =a$

Distance from $(0,0)$

$\to \frac{|a|}{\sqrt{{\sec }^2\theta +cose{c}^2\theta }}=p$ ...(1)

$x\cos \theta +y\sin \theta =a{\cos }^2\theta$

distance from $(0,0)$

$\Rightarrow \frac{|-a{\cos }^2\theta |}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}=q$

$\Rightarrow aco{s}^2\theta =q$ .....(2)

$\Rightarrow {\cos }^{2}2\theta =\frac{q^2}{a^2}$ ......(3)

From equation (1)

$\Rightarrow \frac{a}{p}=\sqrt{\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }}$

$\Rightarrow \frac{a^2}{p^2}=\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }=\frac{1}{{\sin }^2\theta {\cos }^2\theta }$

$\Rightarrow \frac{a^2}{4p^2}=\frac{1}{4{\sin }^2\theta {\cos }^2\theta }$

$\Rightarrow {\sin }^{2}2\theta =\frac{4p^2}{a^2}$ ...(4)

From equation (3) and (4), we get

$\frac{q^2}{a^2}+\frac{4p^2}{a^2}=1$

$4p^2+q^2=a^2$