Question

If $P_1,P_2,P_3$ be the perpendicular from the points $(m^2,2m)$, $(m{m}^{\prime },m+m^{\prime })$ and $(m^2,2m)$ respectively on the line $xcosa+ysina+si{n}^{2}a/co{s}^{2}a=0$, then $P_1,P_2,P_3$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. none of these

Answer 1

The correct option is B G.P.

Given line

$x\cos a+y\sin a+\frac{{\sin }^{2}a}{{\cos }^{2}a}=0---\left(1\right)$

$P_1$ is the perpendicular distance of given line (1) from point $(m^2,2m)$

$P_1=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

$P_1=\left|\frac{m^2\cos a+2m\sin a+\frac{{\sin }^{2}a}{{\cos }^{2}a}}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}}\right|$

$P_1=\left|\frac{m^2{\cos }^{3}a+2m\sin a{\cos }^{2}a+{\sin }^{2}a}{{\cos }^{2}a}\right|$

$P_2$ is the perpendicular distance of given line (1) from point $(m{m}^{\prime },m+m^{\prime })$

$P_2=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

$P_2=\left|\frac{m{m}^{\prime }\cos a+m\sin a+m^{\prime }\sin a+\frac{{\sin }^{2}a}{{\cos }^{2}a}}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}}\right|$

$P_2=\left|\frac{m{m}^{\prime }{\cos }^{3}a+m\sin a{\cos }^{2}a+m^{\prime }\sin a{\cos }^{2}a+{\sin }^{2}a}{{\cos }^{2}a}\right|$

$P_3$ is the perpendicular distance of given line (1) from point $(m^{\prime 2},2m^{\prime })$

$P_3=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

$P_3=\left|\frac{m^{\prime 2}\cos a+m^{\prime }\sin a+\frac{{\sin }^{2}a}{{\cos }^{2}a}}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}}\right|$

$P_3=\left|\frac{m^{\prime 2}{\cos }^{3}a+m^{\prime }\sin a{\cos }^{2}a+{\sin }^{2}a}{{\cos }^{2}a}\right|$

$P_2^2=P_1{P}_2$

Hence $P_1,P_2,P_3$ are in $G.P$