Question

If the chord joining the points $\left( at_{1}^{2},2a{{t}_{1}} \right)$ and $P\left( at_{2}^{2},2a{{t}_{2}} \right)$ of the parabola ${{y}^{2}}=4ax$ passes through the focus of the parabola, then
A.${{t}_{1}}{{t}_{2}}=-1$
B.${{t}_{1}}{{t}_{2}}=1$
C.${{t}_{1}}+{{t}_{2}}=-1$
D.${{t}_{1}}-{{t}_{2}}=1$

Answer 1

Hint: Take P and Q as the coordinates of the focal chord. $S\left( a,o \right)$ will be the focus. Thus PS and SQ has the same slope. Find the slope of PS and QS separately and equate them to get the relation between ${{t}_{1}}$ and ${{t}_{2}}$ .

Complete step-by-step answer:
The cord of the parabola which passes through the focus is called the focal chord. Now we have been given the general equation of parabola as ${{y}^{2}}=4ax$ .

Let us take the points on the parabola as $P\left( at_{1}^{2},2a{{t}_{1}} \right)$ and $Q\left( at_{2}^{2},2a{{t}_{2}} \right)$ .Any chord to ${{y}^{2}}=4ax$ which para through the focus is called focal chord on the parabola ${{y}^{2}}=4ax$.
We said that $P\left( at_{1}^{2},2a{{t}_{1}} \right)$ .The coordinate of the other extreme Q of the focal chord is through P is $\left( at_{2}^{2},2a{{t}_{2}} \right)$ .Then PS and SQ , where $S\left( a,o \right)$ is the focus has the same slope.
I.e. slope of PS = Slope of QS.
We know the formula of slope as $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Let us first find the slope of PS. Take $\left( {{x}_{1}},{{y}_{1}} \right)=\left( a,o \right)$ and ${{x}_{2}},{{y}_{2}}=\left( at_{1}^{2},2a{{t}_{1}} \right)$
Slope of PS $=\dfrac{2a{{t}_{1}}-0}{at_{1}^{2}-a}=\dfrac{2a{{t}_{1}}}{a\left( t_{1}^{2}-1 \right)}=\dfrac{2{{t}_{1}}}{t_{1}^{2}-1}$
$\therefore $ slope of QS $=\dfrac{2{{t}_{2}}}{t_{2}^{2}-1}$ .
Now let us equate both the slopes,
Slope of PS = Slope of QS
$\dfrac{2{{t}_{1}}}{t_{1}^{2}-1}=\dfrac{2{{t}_{2}}}{t_{2}^{2}-1}$ can 2 from each side and cross multiply.
$\begin{align}
  & \therefore {{t}_{1}}\left( t_{2}^{2}-1 \right)={{t}_{2}}\left( t_{1}^{2}-1 \right) \\
 & {{t}_{1}}t_{2}^{2}-{{t}_{1}}={{t}_{2}}t_{1}^{2}-{{t}_{2}} \\
 & \Rightarrow {{t}_{1}}t_{2}^{2}-{{t}_{2}}t_{1}^{2}-{{t}_{1}}+{{t}_{2}}=0 \\
 & {{t}_{1}}{{t}_{2}}\left( {{t}_{2}}-{{t}_{1}} \right)+\left( {{t}_{2}}-{{t}_{1}} \right)=0 \\
 & \Rightarrow \left( {{t}_{1}}.{{t}_{2}}+1 \right)\left( {{t}_{2}}-{{t}_{1}} \right)=0 \\
\end{align}$
Thus from the above equation we can take, ${{t}_{1}}.{{t}_{2}}+1=0$
$\therefore {{t}_{1}}{{t}_{2}}=-1$ i.e. ${{t}_{2}}=\dfrac{-1}{{{t}_{1}}}$ , the point is at $Q\left( \dfrac{a}{{{t}_{2}}},-\dfrac{2a}{t} \right)$
Thus we got the four of the parabola ${{t}_{1}}{{t}_{2}}=-1$ .
$\therefore $ option (A) is the correct answer.
Note: One of the basic identities of the parabola is that if ${{t}_{1}}$ and ${{t}_{2}}$ are the endpoints of a focal chord. Then their slopes can be connected, so that we get ${{t}_{1}}{{t}_{2}}=-1$ . We can answer this question even without the complete solution.