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Question: If \({t_1}\) and \({t_2}\) are two extremities of any focal chord of the parabola \({y^2} = 4ax\) th...

If t1{t_1} and t2{t_2} are two extremities of any focal chord of the parabola y2=4ax{y^2} = 4ax thent1t2={t_1}{t_2}=
A. −1 - 1
B. 00
C. ±1 \pm 1
D. 12\dfrac{1}{2}

Explanation

Solution

Hint: In order to solve the problem just find the slopes with the help of coordinates. Then equate both the slopes and get a relation.
Coordinates of end point of focal chord are (at12,2at1),(at22,2at2)\left( {at_1^2,2a{t_1}} \right),\left( {at_2^2,2a{t_2}} \right)and the focus is(a,0)\left( {a,0} \right)
Three points are collinear, so slopes will be same,
⇒2at2−2at1at22−at12=2at2−0at22−a ⇒t2−t1t22−t12=t2t22−1 ⇒t2−t1(t2−t1)(t2+t1)=t2t22−1 ⇒1(t2+t1)=t2t22−1 ⇒t22−1=t2(t2+t1)=t1t2+t22 ⇒t1t2=−1  \Rightarrow \dfrac{{2a{t_2} - 2a{t_1}}}{{at_2^2 - at_1^2}} = \dfrac{{2a{t_2} - 0}}{{at_2^2 - a}} \\\ \Rightarrow \dfrac{{{t_2} - {t_1}}}{{t_2^2 - t_1^2}} = \dfrac{{{t_2}}}{{t_2^2 - 1}} \\\ \Rightarrow \dfrac{{{t_2} - {t_1}}}{{\left( {{t_2} - {t_1}} \right)\left( {{t_2} + {t_1}} \right)}} = \dfrac{{{t_2}}}{{t_2^2 - 1}} \\\ \Rightarrow \dfrac{1}{{\left( {{t_2} + {t_1}} \right)}} = \dfrac{{{t_2}}}{{t_2^2 - 1}} \\\ \Rightarrow t_2^2 - 1 = {t_2}\left( {{t_2} + {t_1}} \right) = {t_1}{t_2} + t_2^2 \\\ \Rightarrow {t_1}{t_2} = - 1 \\\
Hence option A is the correct option.

Note:In such a question where indirectly something is asked from the question, do not try to find all the points, rather try to manipulate the equation with the help of slopes. Like in this problem the 3 points lie on the same line hence collinearity condition could be used.