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

$ \therefore p_1 p_2=\left|\frac{b \sqrt{a^2-b^2} \cos \theta-a b}{\sqrt{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}}\right| $
$\times\left|\frac{-b \sqrt{a^2-b^2} \cos \theta-a b}{\sqrt{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}}\right| $
$=\frac{b\left|\sqrt{a^2-b^2} \cos \theta-a\right| \times b\left|\sqrt{a^2-b^2} \cos \theta+a\right|}{\left(b^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right)} $
$=\frac{b^2\left|\left(a^2-b^2\right) \cos ^2 \theta-a^2\right|}{\left(b^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right)} $
$\left[\because(A+B)(A-B)=A^2-B^2\right]$
$=\frac{b^2\left|a^2 \cos ^2 \theta-b^2 \cos ^2 \theta-a^2\right|}{\left(b^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right)}$
$=\frac{b^2\left|a^2\left(\cos ^2 \theta-1\right)-b^2 \cos ^2 \theta\right|}{\left(b^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right)} $
$ \Rightarrow p_1 p_2=\frac{b^2\left|-a^2 \sin ^2 \theta-b^2 \cos ^2 \theta\right|}{\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\left|a^2 \sin ^2 \theta+b^2 \cos ^2 \theta\right|}{\left(b^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right)} (\because|-A|=|A|)$
$ \therefore p_1 p_2 =b^2$