Question

If $x\cos \alpha +y\sin \alpha =p$, where $p=(-\frac{{\sin }^2\alpha }{\cos \alpha })$ be a straight line, then perpendiculars on this straight line from the points $(m^2,2m),(m^{\prime }m,m+m^{\prime }),(m^{\prime 2}m,2m^{\prime })$ form a G.P.

• A. True
• B. False

Answer 1

The correct option is A True

Given line

$xcos\alpha +ysin\alpha +\frac{si{n}^2\alpha }{co{s}^2\alpha }=0-\left(1\right)$

​

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

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

$P_1=|\frac{m^{2}cos\alpha +2msin\alpha +\frac{si{n}^2\alpha }{co{s}^2\alpha }}{\sqrt{co{s}^2\alpha +si{n}^2\alpha }}$

$P_1=|\frac{m^{2}co{s}^3\alpha +2msin\alpha co{s}^2\alpha +si{n}^2\alpha }{co{s}^2\alpha }|$

​Similarly,

$P_2$ is the perpendicular distance of given line (1) from point (mm',m+m')

$P_2=|\frac{m{m}^{\prime }co{s}^3\alpha +msin\alpha co{s}^2\alpha +m^{\prime }sin\alpha co{s}^2\alpha }{co{s}^2\alpha }|-\left(2\right)$

Similarly,

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

$P_3=|\frac{m^{\prime 2}co{s}^3\alpha +m^{\prime }sin\alpha co{s}^2\alpha +si{n}^2\alpha }{co{s}^2\alpha }|-\left(3\right)$

From eqs.(1),(2), and (3)

$P_2^2=P_1{P}_2$

Hence $P_1,P_{2}and{P}_3$ are in G.P.