Question

A line which makes an acute angle $\theta$ with the positive direction of the $x$-axis is drawn through the point $P(3,4)$ to meet the line $x=6$ at $R$ and $y=8$ at $S$. Then,

• A. $PR=3\sec \theta$
• B. $PS=4\mathrm{cosec}\theta$
• C. $PR+PS=\frac{2(3\sin \theta +4\cos \theta )}{\sin 2\theta }$
• D. $\frac{9}{(PR{)}^2}+\frac{16}{(PS{)}^2}=1$

Answer 1

Answer: (A), (B), (C), (D)

The equation of any line making an acute angle $\theta$ with the positive direction of the $x$-axis and passing through $P(3,4)$ is $\frac{x-3}{\cos \theta }=\frac{y-4}{\sin \theta }=r$
where $|r|$ is the distance of any point $(x,y)$ from $P$. Therefore, $A(r\cos \theta +3,r\sin \theta +4)$ is a general point on line (i). If $A$ is $R$, then
$r\cos \theta +3=6\text{ or }r=\frac{3}{\cos \theta }=3\sec \theta$
Since $\theta$ is acute, $\cos \theta >0$. Therefore,
$PR=r=3\sec \theta$
If $A\equiv S,r\sin \theta +4=8$. Therefore,
$r=4\mathrm{cosec}\theta$
$\therefore PS=4\mathrm{cosec}\theta$
Also, $PR+PS=\frac{3}{\cos \theta }+\frac{4}{\sin \theta }$
$=\frac{2(3\sin \theta +4\cos \theta )}{\sin 2\theta }$
and $\frac{9}{(PR{)}^2}+\frac{16}{(PS{)}^2}={\cos }^2\theta +{\sin }^2\theta =1$
Therefore, (1), (2), (3), and (4) all are correct.