Question

If ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ ​ be the perpendicular from the points $\left( {{m}^{2}},2m \right),\left( m{m}',m+{m}' \right)$and $\left( {{{{m}'}}^{2}},2{m}' \right)$ respectively on the line $x\cos a+y\sin a+\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}=0$, then ${{P}_{1}},{{P}_{2}},{{P}_{3}}$​ are in
A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer 1

In order to find the solution of the given question that is if ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ ​ be the perpendicular from the points $\left( {{m}^{2}},2m \right),\left( m{m}',m+{m}' \right)$and $\left( {{{{m}'}}^{2}},2{m}' \right)$ respectively on the line $x\cos a+y\sin a+\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}=0$, then ${{P}_{1}},{{P}_{2}},{{P}_{3}}$​ are in A.P./ G.P./H.P./ or none of these. Apply the formula of the perpendicular distance of a line $ax+by+c$ from the point $\left( x,y \right)$ is $d=\left| \dfrac{ax+by+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$. Now use this formula to find the perpendicular distances ${{P}_{1}}$​: given by line $x\cos a+y\sin a+\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}=0$ from point $\left( {{m}^{2}},2m \right)$; ${{P}_{2}}$: given by line $x\cos a+y\sin a+\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}=0$ from point $\left( m{m}',m+{m}' \right)$; ${{P}_{3}}$: given by line $x\cos a+y\sin a+\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}=0$ from point $\left( {{{{m}'}}^{2}},2{m}' \right)$. After this check for the conditions if distance ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ are in A.P. then the difference between any two consecutive terms will be constant. Then check for the conditions if distance ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ are in G.P. then any element after the first is obtained by multiplying the preceding element by a constant. To check for the conditions if distance ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ are in H.P. then the reciprocals of the terms of a sequence are in arithmetic progression.

Complete step by step solution:
According to the question, given line in the question is as follows:
$x\cos a+y\sin a+\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}=0...\left( 1 \right)$
We know that the perpendicular distance of a line $ax+by+c$ from the point $\left( x,y \right)$ is $d=\left| \dfrac{ax+by+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$.
Therefore, ${{P}_{1}}$​ is the perpendicular distance of given line $\left( 1 \right)$ from point $\left( {{m}^{2}},2m \right)$ is as follows:

${{P}_{1}}=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$
$\Rightarrow {{P}_{1}}=\left| \dfrac{{{m}^{2}}\cos a+2m\sin a+\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}}{\sqrt{{{\cos }^{2}}a+{{\sin }^{2}}a}} \right|$
$\Rightarrow {{P}_{1}}=\left| \dfrac{{{m}^{2}}{{\cos }^{3}}a+2m\sin a{{\cos }^{2}}a+{{\sin }^{2}}a}{\sqrt{{{\cos }^{2}}a+{{\sin }^{2}}a}\left( {{\cos }^{2}}a \right)} \right|$
$\Rightarrow {{P}_{1}}=\left| \dfrac{{{m}^{2}}{{\cos }^{3}}a+2m\sin a{{\cos }^{2}}a+{{\sin }^{2}}a}{\sqrt{1}\left( {{\cos }^{2}}a \right)} \right|$
$\Rightarrow {{P}_{1}}=\left| \dfrac{{{m}^{2}}{{\cos }^{3}}a+2m\sin a{{\cos }^{2}}a+{{\sin }^{2}}a}{{{\cos }^{2}}a} \right|...\left( 2 \right)$
We know that the perpendicular distance of a line $ax+by+c$ from the point $\left( x,y \right)$ is $d=\left| \dfrac{ax+by+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$.
Therefore,${{P}_{2}}$​ is the perpendicular distance of given line $\left( 1 \right)$ from point $\left( m{m}',m+{m}' \right)$
${{P}_{2}}=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$
$\Rightarrow {{P}_{2}}=\left| \dfrac{m{m}'\cos a+m\sin a+{m}'\sin a+\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}}{\sqrt{{{\cos }^{2}}a+{{\sin }^{2}}a}} \right|$
$\Rightarrow {{P}_{2}}=\left| \dfrac{m{m}'{{\cos }^{3}}a+m\sin a{{\cos }^{2}}a+{m}'\sin a{{\cos }^{2}}a+{{\sin }^{2}}a}{\sqrt{{{\cos }^{2}}a+{{\sin }^{2}}a}\left( {{\cos }^{2}}a \right)} \right|$
$\Rightarrow {{P}_{2}}=\left| \dfrac{m{m}'{{\cos }^{3}}a+m\sin a{{\cos }^{2}}a+{m}'\sin a{{\cos }^{2}}a+{{\sin }^{2}}a}{\sqrt{1}\left( {{\cos }^{2}}a \right)} \right|$
$\Rightarrow {{P}_{2}}=\left| \dfrac{m{m}'{{\cos }^{3}}a+m\sin a{{\cos }^{2}}a+{m}'\sin a{{\cos }^{2}}a+{{\sin }^{2}}a}{{{\cos }^{2}}a} \right|...\left( 3 \right)$
We know that the perpendicular distance of a line $ax+by+c$ from the point $\left( x,y \right)$ is $d=\left| \dfrac{ax+by+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$.
Therefore,${{P}_{3}}$​ is the perpendicular distance of given line $\left( 1 \right)$ from point $\left( {{{{m}'}}^{2}},2{m}' \right)$
${{P}_{3}}=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$
$\Rightarrow {{P}_{3}}=\left| \dfrac{{{{{m}'}}^{2}}\cos a+2{m}'\sin a+\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}}{\sqrt{{{\cos }^{2}}a+{{\sin }^{2}}a}} \right|$
$\Rightarrow {{P}_{3}}=\left| \dfrac{{{{{m}'}}^{2}}{{\cos }^{3}}a+2{m}'\sin a{{\cos }^{2}}a+{{\sin }^{2}}a}{\sqrt{{{\cos }^{2}}a+{{\sin }^{2}}a}\left( {{\cos }^{2}}a \right)} \right|$
$\Rightarrow {{P}_{3}}=\left| \dfrac{{{{{m}'}}^{2}}{{\cos }^{3}}a+2{m}'\sin a{{\cos }^{2}}a+{{\sin }^{2}}a}{\sqrt{1}\left( {{\cos }^{2}}a \right)} \right|$
$\Rightarrow {{P}_{3}}=\left| \dfrac{{{{{m}'}}^{2}}{{\cos }^{3}}a+2{m}'\sin a{{\cos }^{2}}a+{{\sin }^{2}}a}{{{\cos }^{2}}a} \right|...\left( 4 \right)$
Now, checking for the conditions if distance ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ are in A.P. then the difference between any two consecutive terms will be constant. But this does not satisfy with ${{P}_{1}},{{P}_{2}},{{P}_{3}}$. Therefore, ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ are not in A.P.
Now check for the conditions if distance ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ are in G.P. then any element after the first is obtained by multiplying the preceding element by a constant.
We can clearly see from the equation $\left( 2 \right)$, $\left( 3 \right)$ and $\left( 4 \right)$, we get:
${{P}_{2}}^{2}={{P}_{1}}{{P}_{3}}$
Hence, we can write ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ in G.P. form.
Therefore, ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ are in G.P.

So, the correct answer is “Option B”.

Note: Students get confused between the concepts of A.P. and G.P., its important to remember that a sequence is in A.P. if the difference between any two consecutive terms will be constant and a sequence is in G.P. then any element after the first is obtained by multiplying the preceding element by a constant.