Question

If l and l’ are the lengths of segment of focal chord of a parabola ${{y}^{2}}=4ax$, then prove that $\dfrac{1}{l}+\dfrac{1}{l'}=\dfrac{1}{a}$.

Answer 1

Hint: In the above question we will use the parametric form of the parabola ${{y}^{2}}=4ax$ which is $\left( a{{t}^{2}},2at \right)$. We will use the property of a focal chords, if one end of a focal chord A $=\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right)$ and the other end of a focal chord B $=\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right)$ then ${{t}_{1}}{{t}_{2}}=-1$.

Complete step-by-step answer:
We will also use the distance formula between two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ as follows:
Distance $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

We have been given a parabola ${{y}^{2}}=4ax$ and l and l’ are the lengths of segment of focal chord of parabola and then we have to prove $\dfrac{1}{l}+\dfrac{1}{l'}=\dfrac{1}{a}$.
We know the parametric form of the parabola ${{y}^{2}}=4ax$ is $\left( a{{t}^{2}},2at \right)$.
Let us suppose the end points of focal chord A $\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right)$ and the other end point B $\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right)$.
By the property of focal chords we know that ${{t}_{1}}{{t}_{2}}=-1$
$\Rightarrow {{t}_{2}}=\dfrac{-1}{{{t}_{1}}}$
On substituting the value of ${{t}_{2}}$ in point B, we get as follows:

& a{{t}_{2}}^{2}=a{{\left( \dfrac{-1}{{{t}_{1}}} \right)}^{2}}=\dfrac{a}{{{t}_{1}}^{2}} \\\ & 2a{{t}_{2}}=2a\left( \dfrac{-1}{{{t}_{1}}} \right)=\dfrac{-2a}{{{t}_{1}}} \\\ \end{aligned}$$ So the point B is $$\left( \dfrac{a}{{{t}_{1}}^{2}},\dfrac{-2a}{{{t}_{1}}} \right)$$. Also, focus is S = (a,0)  We know the distance formula between two points $$\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$\left( {{x}_{2}},{{y}_{2}} \right)$$ as follows: Distance $$=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$$ By using the distance formula the length of segment of focal chords are as follows: $$l=AS=\sqrt{{{\left( a{{t}_{1}}^{2}-a \right)}^{2}}+{{\left( 2a{{t}_{1}} \right)}^{2}}}$$ Using the identity $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$ in the above equation, we get as follows: $$\begin{aligned} & \Rightarrow \sqrt{{{a}^{2}}{{\left( {{t}_{1}}^{2}-1 \right)}^{2}}+4{{a}^{2}}{{t}_{1}}^{2}} \\\ & \Rightarrow \sqrt{{{a}^{2}}{{t}_{1}}^{4}-2{{a}^{2}}{{t}_{1}}^{2}+{{a}^{2}}+4{{a}^{2}}{{t}_{1}}^{2}} \\\ & \Rightarrow \sqrt{{{a}^{2}}{{t}_{1}}^{4}+2{{a}^{2}}{{t}_{1}}^{2}+{{a}^{2}}} \\\ & \Rightarrow \sqrt{{{a}^{2}}{{\left( {{t}_{1}}^{2}+1 \right)}^{2}}} \\\ & \Rightarrow a\left( {{t}_{1}}^{2}+1 \right) \\\ \end{aligned}$$ $$l'=BS=\sqrt{{{\left( \dfrac{a}{{{t}_{1}}^{2}}-a \right)}^{2}}+{{\left( \dfrac{-2a}{{{t}_{1}}}-0 \right)}^{2}}}$$ Using the identity $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$ in the above equation, we get as follows: $$\begin{aligned} & \Rightarrow \sqrt{\dfrac{{{a}^{2}}}{{{t}_{1}}^{4}}-\dfrac{2{{a}^{2}}}{{{t}_{1}}^{2}}+{{a}^{2}}+\dfrac{4{{a}^{2}}}{{{t}_{1}}^{2}}} \\\ & \Rightarrow \sqrt{\dfrac{{{a}^{2}}}{{{t}_{1}}^{4}}+\dfrac{2{{a}^{2}}}{{{t}_{1}}^{2}}+{{a}^{2}}} \\\ & \Rightarrow \sqrt{{{\left( \dfrac{a}{{{t}_{1}}^{2}}+a \right)}^{2}}} \\\ & \Rightarrow \dfrac{a}{{{t}_{1}}^{2}}+a=a\left( \dfrac{1}{{{t}_{1}}^{2}}+1 \right) \\\ & \Rightarrow \dfrac{a\left( 1+{{t}_{1}}^{2} \right)}{{{t}_{1}}^{2}} \\\ \end{aligned}$$ Now, $$\dfrac{1}{l}+\dfrac{1}{l'}=\dfrac{1}{a\left( {{t}_{1}}^{2}+1 \right)}+\dfrac{{{t}_{1}}^{2}}{a\left( {{t}_{1}}^{2}+1 \right)}$$ Taking $$\dfrac{1}{a\left( {{t}_{1}}^{2}+1 \right)}$$ as common we get as follows: $$\Rightarrow \dfrac{1}{a\left( {{t}_{1}}^{2}+1 \right)}\left[ 1+{{t}_{1}}^{2} \right]=\dfrac{1}{a}$$ Hence, it is proved that $$\dfrac{1}{l}+\dfrac{1}{l'}=\dfrac{1}{a}$$. Note: Be careful while doing calculation as there is a chance that you might make a sign mistake while using the distance formula as well as finding the value of ‘$${{t}_{2}}$$’ in terms of ‘$${{t}_{1}}$$’ by using the property of focal chord. Also remember that a focal chord means the chord that passes through the focus of the parabola.