Question

If $P_1$ and $P_2$ are the lengths of the perpendiculars drawn from the points $\left(\sqrt{a^2-b^2},0\right)$ and $\left(-\sqrt{a^2-b^2},0\right)$ respectively, to the line $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ then

• A. $P_1{P}_2=b^3$
• B. $P_1{P}_2=b^2$
• C. Both(a) and (b) are true
• D. Neither(a) nor (b) are true

Answer 1

Answer: (B)

Given, equation of line is
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$
$\Rightarrow \frac{bx\cos \theta +ay\sin \theta }{ab}=1$
$\Rightarrow bx\cos \theta +ay\sin \theta =ab$
$\Rightarrow bx\cos \theta +ay\sin \theta -ab=\mathrm{0.....}$(i)
At point $\left(\sqrt{a^2-b^2},0\right)$, the perpendicular distance from line (i) is
$p_1=\left|\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}\right|$
$=\left|\frac{b\sqrt{a^2-b^2}\cos \theta -0-ab}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}\right|$
$\left(\because A=b\cos \theta ,B=a\sin \theta ,C=-ab;x_1=\sqrt{a^2-b^2},y_1=0\right)$
and at point $\left(-\sqrt{a^2-b^2},0\right)$, the perpendicular distance from line (i) is
$p_2=\left|\frac{-b\sqrt{a^2-b^2}\cos \theta -ab}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}\right|$
$\left(\because A=b\cos \theta ,B=a\sin \theta ,C=-ab,x_1=-\sqrt{a^2-b^2},y_1=0\right)$

$\therefore p_1{p}_2=\left|\frac{b\sqrt{a^2-b^2}\cos \theta -ab}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}\right|$
$\times \left|\frac{-b\sqrt{a^2-b^2}\cos \theta -ab}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}\right|$
$=\frac{b|\sqrt{a^2-b^2}\cos \theta -a|\times b|\sqrt{a^2-b^2}\cos \theta +a|}{\left(b^2{\cos }^2\theta +a^2{\sin }^2\theta \right)}$
$=\frac{b^2|\left(a^2-b^2\right){\cos }^2\theta -a^2|}{\left(b^2{\cos }^2\theta +a^2{\sin }^2\theta \right)}$
$\left[\because (A+B)(A-B)=A^2-B^2\right]$
$=\frac{b^2|a^2{\cos }^2\theta -b^2{\cos }^2\theta -a^2|}{\left(b^2{\cos }^2\theta +a^2{\sin }^2\theta \right)}$
$=\frac{b^2|a^2\left({\cos }^2\theta -1\right)-b^2{\cos }^2\theta |}{\left(b^2{\cos }^2\theta +a^2{\sin }^2\theta \right)}$
$\Rightarrow p_1{p}_2=\frac{b^2|-a^2{\sin }^2\theta -b^2{\cos }^2\theta |}{\left(b^2{\cos }^2\theta +a^2{\sin }^2\theta \right)}$
$\left(\because {\sin }^2\theta +{\cos }^2\theta =1\right)$
$=\frac{b^2|a^2{\sin }^2\theta +b^2{\cos }^2\theta |}{\left(b^2{\cos }^2\theta +a^2{\sin }^2\theta \right)}(\because |-A|=|A|)$
$\therefore p_1{p}_2=b^2$