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

Hint: 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.