Question

If t is the parameter for one end of a focal chord of the parabola ${{y}^{2}}=4ax$ , then its length is
(a).$a{{\left( t+\dfrac{1}{t} \right)}^{2}}$
(b).$a{{\left( t-\dfrac{1}{t} \right)}^{2}}$
(c).$a\left( t+\dfrac{1}{t} \right)$
(d).$a\left( t-\dfrac{1}{t} \right)$

Answer 1

Hint: Evaluate the parametric coordinates of the focal chord of parabola ${{y}^{2}}=4ax$ where the parameter is t, and then apply the distance formula to determine the length of the focal chord in terms of t.

Complete step-by-step answer:

Any chord to ${{y}^{2}}=4ax$ which passes through the focus is called a focal chord of the parabola ${{y}^{2}}=4ax$. Focus can be defined as a point in parabola with coordinates $\left( a,0 \right)$. Consider a point P on the parabola whose coordinate in parametric form be $\left( a{{t}^{2}},2at \right)$. For the other extremity Q of the focal chord through P, the coordinates of Q be $\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right)$.

To establish a relationship between t and ${{t}_{1}}$ we use the expression of slope of a straight line. Now, consider the coordinates of two points be $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$.

The slope of the line joining these points would be, $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.

Let PS be the part of the focal chord above the x-axis and QS be the part of the focal chord below the x-axis.

Then, PS and SQ, where S is the focus $\left( a,0 \right)$, have the same slopes because it represents one single straight line.
$\begin{aligned} & \Rightarrow \dfrac{2at-0}{a{{t}^{2}}-a}=\dfrac{2a{{t}_{1}}-0}{a{{t}_{1}}^{2}-a} \\\ & t{{t}_{1}}^{2}-t={{t}_{1}}{{t}^{2}}-{{t}_{1}} \\\ & \left( t{{t}_{1}}+1 \right)\left( {{t}_{1}}-1 \right)=0 \\\ & \therefore {{t}_{1}}=\dfrac{-1}{t} \\\ \end{aligned}$

Therefore, the point Q is $\left( \dfrac{a}{{{t}^{2}}},\dfrac{-2a}{t} \right)$ .

Now, to find the distance between two points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$, we employ the distance formula which can be stated as:
$D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
From the above steps, the two ends of the focal chord are evaluated as:
$P=\left( a{{t}^{2}},2at \right)\text{ and }Q=\left( \dfrac{a}{{{t}^{2}}},-\dfrac{2a}{t} \right)$

Finally, we are required to evaluate the distance PQ, by putting the coordinates of P and Q in the distance formula.
So, the focal length is:
$\begin{aligned} & PQ=\sqrt{{{\left( a{{t}^{2}}-\dfrac{a}{{{t}^{2}}} \right)}^{2}}+{{\left( 2at-\left( -\dfrac{2a}{t} \right) \right)}^{2}}} \\\ & PQ=\sqrt{{{\left( a{{t}^{2}}-\dfrac{a}{{{t}^{2}}} \right)}^{2}}+{{\left( 2at+\dfrac{2a}{t} \right)}^{2}}} \\\ \end{aligned}$

Taking out a so that we get simplified expression,
$\begin{aligned} & PQ=a\sqrt{{{\left( {{t}^{2}}-\dfrac{1}{{{t}^{2}}} \right)}^{2}}+4{{\left( t+\dfrac{1}{t} \right)}^{2}}} \\\ & PQ=a\sqrt{{{\left( t+\dfrac{1}{t} \right)}^{2}}{{\left( t-\dfrac{1}{t} \right)}^{2}}+4{{\left( t+\dfrac{1}{t} \right)}^{2}}} \\\ \end{aligned}$

Now taking $\left( t+\dfrac{1}{t} \right)$ common,
$\begin{aligned} & PQ=a\left( t+\dfrac{1}{t} \right)\sqrt{{{\left( t-\dfrac{1}{t} \right)}^{2}}+4} \\\ & PQ=a\left( t+\dfrac{1}{t} \right)\sqrt{{{t}^{2}}+\dfrac{1}{{{t}^{2}}}-2+4} \\\ & PQ=a\left( t+\dfrac{1}{t} \right)\sqrt{{{t}^{2}}+\dfrac{1}{{{t}^{2}}}+2} \\\ & PQ=a\left( t+\dfrac{1}{t} \right)\sqrt{{{\left( t+\dfrac{1}{t} \right)}^{2}}} \\\ & PQ=a{{\left( t+\dfrac{1}{t} \right)}^{2}} \\\ \end{aligned}$

Hence, the length of the focal chord is $PQ=a{{\left( t+\dfrac{1}{t} \right)}^{2}}$.

Therefore, the correct option is (a).

Note: The key step in solving this problem is the expression of coordinate of focal chord in terms of parameter t.
One common confusion is which equation is valid for establishing the link between t and ${{t}_{1}}$, as we got two expressions in the slope relationship.
So, we would consider that equation which expresses one parameter in the form of another and discard the other relation.