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Question

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(a) A.P

(b) G.P

(c) H.P

(d) None of these

Answer

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Hint: Here we use the equation of parabola and find the perpendicular distance from A, B, C on tangent to a curve using distance formula.

Complete step-by-step answer:

The given equation of parabola is ${y^2} = 4ax$

Let us suppose that $B\left( {at_1^2,2a{t_1}} \right)$ and $C\left( {at_2^2,2a{t_2}} \right)$ be general points on the curve

Now according to parametric form the point from where tangents are drawn to the parabola that is A will have coordinates $\left( {a{t_1}{t_2},a\left( {{t_1} + {t_2}} \right)} \right)$ equation of the tangent can be written as $y = mx + \dfrac{a}{m}$ where y and x are coordinates respectively and m is slope and a is foci.

In question it is mentioned that ${p_2}$ is the length of perpendicular drawn from B point to any of the tangent on this curve ${y^2} = 4ax$

So the perpendicular distance from a point (m, n) to a line \[Ax + By + C = 0\] is given by

\[d = \left| {\dfrac{{Am + Bn + C}}{{\sqrt {{A^2} + {B^2}} }}} \right|\] $\to$ (1)

Now using the concept from equation (1) we can write,

${p_2} = \left| {\dfrac{{2a{t_1} - mat_1^2 - \dfrac{a}{m}}}{{\sqrt {1 + {m^2}} }}} \right|$

Now taking $\dfrac{a}{m}$ common we get

\[\left| {\dfrac{a}{m}} \right|\dfrac{{{{\left( {m{t_1} - 1} \right)}^2}}}{{\sqrt {1 + {m^2}} }} = {p_1}\] $\to$ (2)

Similarly ${p_3}$ is the length of perpendicular drawn from point C to any tangent on the given curve using the concept of (1) our ${p_3}$ will be equal to

${p_3} = \left| {\dfrac{{2a{t_2} - mat_2^2 - \dfrac{a}{m}}}{{\sqrt {1 + {m^2}} }}} \right|$

Now taking $\dfrac{a}{m}$ common we get

$\left| {\dfrac{a}{m}} \right|\dfrac{{{{\left( {m{t_2} - 1} \right)}^2}}}{{\sqrt {1 + {m^2}} }} = {p_3}$ $\to$ (3)

Similarly ${p_1}$ is the length of perpendicular drawn from point A to any tangent on the given curve using the concept of (1) our ${p_1}$ will be equal to

${p_1} = \left| {\dfrac{{ma{t_1}{t_2} - a\left( {{t_1} + {t_2}} \right) + \dfrac{a}{m}}}{{\sqrt {1 + {m^2}} }}} \right|$

Now taking $\dfrac{a}{m}$ common we get

$\left| {\dfrac{a}{m}} \right|\dfrac{{\left( {m{t_1} - 1} \right)\left( {m{t_2} - 1} \right)}}{{\sqrt {1 + {m^2}} }} = {p_1}$ $\to$ (4)

Clearly from equation 2, 3, 4 we can see that

$p_1^2 = {p_2}{p_3}$

Now, if we have, ${b^2} = ac$ then we can say that a, b, c are in GP.

Similarly here we can say that ${p_2}, {p_1} \& {p_3}$ are in GP.

Correct answer - Option (b)

Note: Whenever you are given some unknown points on parabola you can assume the parametric form of points that is in general $\left( {a{t^2},2at} \right)$, the concept of perpendicular with tangent can always be solved using the formula mentioned in (1), this parametric point concept from a wide variety of question in JEE so the parametric points should be remembered.

Complete step-by-step answer:

The given equation of parabola is ${y^2} = 4ax$

Let us suppose that $B\left( {at_1^2,2a{t_1}} \right)$ and $C\left( {at_2^2,2a{t_2}} \right)$ be general points on the curve

Now according to parametric form the point from where tangents are drawn to the parabola that is A will have coordinates $\left( {a{t_1}{t_2},a\left( {{t_1} + {t_2}} \right)} \right)$ equation of the tangent can be written as $y = mx + \dfrac{a}{m}$ where y and x are coordinates respectively and m is slope and a is foci.

In question it is mentioned that ${p_2}$ is the length of perpendicular drawn from B point to any of the tangent on this curve ${y^2} = 4ax$

So the perpendicular distance from a point (m, n) to a line \[Ax + By + C = 0\] is given by

\[d = \left| {\dfrac{{Am + Bn + C}}{{\sqrt {{A^2} + {B^2}} }}} \right|\] $\to$ (1)

Now using the concept from equation (1) we can write,

${p_2} = \left| {\dfrac{{2a{t_1} - mat_1^2 - \dfrac{a}{m}}}{{\sqrt {1 + {m^2}} }}} \right|$

Now taking $\dfrac{a}{m}$ common we get

\[\left| {\dfrac{a}{m}} \right|\dfrac{{{{\left( {m{t_1} - 1} \right)}^2}}}{{\sqrt {1 + {m^2}} }} = {p_1}\] $\to$ (2)

Similarly ${p_3}$ is the length of perpendicular drawn from point C to any tangent on the given curve using the concept of (1) our ${p_3}$ will be equal to

${p_3} = \left| {\dfrac{{2a{t_2} - mat_2^2 - \dfrac{a}{m}}}{{\sqrt {1 + {m^2}} }}} \right|$

Now taking $\dfrac{a}{m}$ common we get

$\left| {\dfrac{a}{m}} \right|\dfrac{{{{\left( {m{t_2} - 1} \right)}^2}}}{{\sqrt {1 + {m^2}} }} = {p_3}$ $\to$ (3)

Similarly ${p_1}$ is the length of perpendicular drawn from point A to any tangent on the given curve using the concept of (1) our ${p_1}$ will be equal to

${p_1} = \left| {\dfrac{{ma{t_1}{t_2} - a\left( {{t_1} + {t_2}} \right) + \dfrac{a}{m}}}{{\sqrt {1 + {m^2}} }}} \right|$

Now taking $\dfrac{a}{m}$ common we get

$\left| {\dfrac{a}{m}} \right|\dfrac{{\left( {m{t_1} - 1} \right)\left( {m{t_2} - 1} \right)}}{{\sqrt {1 + {m^2}} }} = {p_1}$ $\to$ (4)

Clearly from equation 2, 3, 4 we can see that

$p_1^2 = {p_2}{p_3}$

Now, if we have, ${b^2} = ac$ then we can say that a, b, c are in GP.

Similarly here we can say that ${p_2}, {p_1} \& {p_3}$ are in GP.

Correct answer - Option (b)

Note: Whenever you are given some unknown points on parabola you can assume the parametric form of points that is in general $\left( {a{t^2},2at} \right)$, the concept of perpendicular with tangent can always be solved using the formula mentioned in (1), this parametric point concept from a wide variety of question in JEE so the parametric points should be remembered.