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More # If the point P on the parabola ${y^2} = 4ax$ for which |PR - PQ| is maximum, where R =(- a,0), Q =(0,a) is represented as $(k_1a, k_2a)$, then find the value of $(k_1+k_2)$.

Last updated date: 13th Mar 2023
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Hint: Use triangle inequality to find the general points of parabola in parametric form to find the sum of $K_1$ and $K_2$.

First draw the diagram of the parabola equation which is horizontally parabola opening right side.

P point is given as $({K_1}a,{K_2}a)$,Q point is given as $(0,a)$ and R point is given as $(- a,0)$

When you join these points it makes a triangle, So according to triangle inequality the modulus of difference of two sides is always less than or equal to the third side.

$\Rightarrow \left| {PR - PQ} \right| \leqslant \left| {QR} \right|$

Maximum value of $\left| {PR - PQ} \right|\; = QR$ because everything is less than or equal to QR.

$\Rightarrow \left| {PR - PQ} \right|\; = QR$, when P, Q, R are collinear

So, the equation of line joining R and Q is

$\Rightarrow y - 0 = \dfrac{{a - 0}}{{0 - \left( { - a} \right)}}\left( {x - \left( { - a} \right)} \right)$

$\Rightarrow y = x + a$

Equation of parabola is $y^2 = 4ax$

In parametric form the coordinates of parabola is $( {a{t^2},2at})$

Where t is a parameter and these parametric coordinates represents every point on parabola.

$\Rightarrow$ p point is $( {a{t^2},2at})$

Now P, Q, R is in straight line, then this p point satisfy the equation of straight line

$\Rightarrow 2at - a{{\text{t}}^2} = a$

$\Rightarrow {t^2} - 2t + 1 = 0$

$\Rightarrow {\left( {t - 1} \right)^2} = 0$

$\Rightarrow \left( {t - 1} \right) = 0$

$\Rightarrow t = 1$

So, put the value of t in P points

$\Rightarrow {\text{P}}\left( {a,2a} \right)$, So compare this points with given points $({K_1}a,{K_2}a)$

So, on comparing $K_1 = 1, K_2 = 2$

$\Rightarrow$ The value of $\left( {{{\text{K}}_1} + {{\text{K}}_2}} \right) = 1 + 2 = 3$