Question

If $p_1,p_2,p_3$ are lengths of perpendiculars from points $(m^2,2m)$ , $(m{m}^{\prime },m+m^{\prime })$ and $({m^{\prime }}^2,2m^{\prime })$ to the line $x\cos \alpha +y\sin \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }=0$ then $p_1,p_2,p_3$ are in

• A. A.P.
• B. G..P.
• C. H.P.
• D. A.G..P.

Answer 1

Answer: (B)

G..P.

$p_1=|m^2\cos \alpha +2m\sin \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }|$
$\Rightarrow p_1=\frac{(m\cos \alpha +\sin \alpha {)}^2}{\cos \alpha }$ ---------(1)
$p_2=|m{m}^{\prime }\cos \alpha +(m+m^{\prime })\sin \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }|$
$p_2=\frac{(m{m}^{\prime }{\cos }^2\alpha +(m+m^{\prime })\sin \alpha \cos \alpha +{\sin }^2\alpha )}{\cos \alpha }$ ------------( 2)
$p_3=|{m^{\prime }}^2\cos \alpha +2m^{\prime }\sin \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }|$
$\Rightarrow p_3=\frac{(m^{\prime }\cos \alpha +\sin \alpha {)}^2}{\cos \alpha }$ ------------(3)
Now taking square root of equation 1 & 3 and multiplying them,

$\sqrt{p_1}\sqrt{p_3}=\frac{(m\cos \alpha +\sin \alpha )}{\sqrt{\cos \alpha }}\frac{(m^{\prime }\cos \alpha +\sin \alpha )}{\sqrt{\cos \alpha }}$
$\Rightarrow \sqrt{p_1}\sqrt{p_3}=\frac{(m{m}^{\prime }{\cos }^2\alpha +(m+m^{\prime })\sin \alpha \cos \alpha +{\sin }^2\alpha )}{\cos \alpha }$

$\Rightarrow p_1{p}_3={p_2}^2$
Hence, $p_1,p_2,p_3$ are in G.P.