Question

If $p_1,p_2,p_3$ be the perpendicular from the points $({\alpha }^2,2\alpha )(\alpha \beta ,\alpha +\beta )$ and
$({\beta }^2,2\beta )$ respectively on the line $xcos\theta +ysin\theta +\frac{{\sin }^2\theta }{\cos \theta }=0$ ,then $p_1,p_2,p_3$ are in:

• A. A.P.
• B. G.P.
• C. H.P.
• D. none of the above

Answer 1

Answer: (B)

G.P.

Using distance formula $\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$
$\begin{array}{c}{P}_1=\frac{{\alpha }^2\cos \theta +2\alpha \sin \theta +\frac{{\sin }^2\theta }{\cos \theta }}{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}\\ ={\alpha }^2\cos \theta +2\alpha \sin \theta +\frac{{\sin }^2\theta }{\cos \theta }\\ =\frac{{\alpha }^2{\cos }^2\theta +2\alpha \sin \theta \cos \theta +{\sin }^2\theta }{\cos \theta }\\ =\frac{{\left(x\cos \theta +sin\theta \right)}^2}{\cos \theta }......\left(1\right)\end{array}$
similarly,
$P_3=\frac{{\left(\beta \cos \theta +sin\theta \right)}^2}{\cos \theta }........\left(2\right)$
And,
$\begin{array}{c}{P}_2=\frac{\alpha \beta \cos \theta +\left(\alpha +\beta \right)\sin \theta +\frac{{\sin }^2\theta }{\cos \theta }}{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}\\ =\frac{\alpha \beta {\cos }^2\theta +\alpha \sin \theta \cos \theta +\beta \sin \theta \cos \theta +{\sin }^2\theta }{\cos \theta }\\ =\frac{\alpha \cos \theta \left[\beta \cos \theta +\sin \theta \right]+\sin \theta \left[\beta \cos \theta +\sin \theta \right]}{\cos \theta }\\ =\frac{\left(\alpha \cos \theta +\sin \theta \right)\left(\beta \cos \theta +\sin \theta \right)}{\cos \theta }........\left(3\right)\end{array}$
From $\left(1\right),\left(2\right)$ and $\left(3\right)$
$\begin{array}{c}{P}_1\times P_3=P_2^2\\ P_1,P_2,P_3\leftarrow GP\end{array}$