Question

The point P on the parabola ${{y}^{2}}=4ax$ for which $\left| PR-PQ \right|$ is maximum, where $R\left( -a,\text{ }0 \right)$ and $Q\text{ }\left( 0,\text{ }a \right)$ is,
A. $\left( a,2a \right)$
B. $\left( a,-2a \right)$
C. $\left( 4a,4a \right)$
D. $\left( 4a,-4a \right)$

Answer 1

Hint: Firstly, draw a diagram to represent the coordinates and the parabola and then decide which length will be denoting the maximum length of $\left| PR-PQ \right|$ . This is possible when PQR is a line. Now equate the slopes of PQ and PR. Use the general denotation for any point on the parabola and then find the coordinate of P.

Complete step-by-step solution:
The given equation of the parabola is ${{y}^{2}}=4ax$
Let P be any point on the parabola.
Any point on a parabola is generally denoted by the coordinates,
$P\left( a{{t}^{2}},2at \right)$
Its diagrammatic representation is given by,

When we look at the triangle PQR,
We can say that always,
$\Rightarrow \left| PR-PQ \right|\le QR$
Hence the maximum value of $\left| PR-PQ \right|$ will be QR.
The maximum value of $\left| PR-PQ \right|$ is QR and is achieved when PQR is a straight line.
And for PQR to be a straight line,
The slope of PR must be equal to the slope of PQ.
Now let us write the slope form for the above condition.
We know that $P\left( a{{t}^{2}},2at \right)$ ,
And $R\left( -a,\text{ }0 \right)$ and $Q\text{ }\left( 0,\text{ }a \right)$
Upon substituting in slope equations, we get,
$\Rightarrow \left( \dfrac{2at-0}{a{{t}^{2}}+a} \right)=\left( \dfrac{2at-a}{a{{t}^{2}}-0} \right)$
Now evaluate to find the value of t from the expression.
$\Rightarrow \left( \dfrac{2at}{a{{t}^{2}}+a} \right)=\left( \dfrac{2at-a}{a{{t}^{2}}} \right)$
Cancel ‘a’ from both sides as it is common on the numerator and denominator.
$\Rightarrow \left( \dfrac{2t}{{{t}^{2}}+1} \right)=\left( \dfrac{2t-1}{{{t}^{2}}} \right)$
Now cross multiply the expressions.
$\Rightarrow 2t\times {{t}^{2}}=\left( {{t}^{2}}+1 \right)\left( 2t-1 \right)$
Evaluate further.
$\Rightarrow 2{{t}^{3}}=2{{t}^{3}}-{{t}^{2}}+2t-1$
Now, cancel the common terms on both sides of the expression.
$\Rightarrow {{t}^{2}}-2t+1=0$
$\Rightarrow {{\left( t-1 \right)}^{2}}=0$
Which upon solving gives us $t=1$
Now substitute this in the general coordinates of $P\left( a{{t}^{2}},2at \right)$ ,
We get,
$\Rightarrow P\left( a{{\left( 1 \right)}^{2}},2a\left( 1 \right) \right)$
$\Rightarrow P\left( a,2a \right)$
Hence option A is correct.

Note: The slope of a line is the steepness of a line in a horizontal or vertical direction. The slope of a line can be calculated by taking the ratio of the change in vertical dimensions upon the change in horizontal dimensions.