If ${t_1}$ and ${t_2}$ are two extremities of any focal chord of the parabola ${y^2} = 4ax$ then${t_1}{t_2}= $
A. $ - 1$
B. $0$
C. $ \pm 1$
D. $\dfrac{1}{2}$
Last updated date: 19th Mar 2023
•
Total views: 310.2k
•
Views today: 6.89k
Answer
310.2k+ views
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 $\left( {at_1^2,2a{t_1}} \right),\left( {at_2^2,2a{t_2}}
\right)$and the focus is$\left( {a,0} \right)$
Three points are collinear, so slopes will be same,
$
\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.
Coordinates of end point of focal chord are $\left( {at_1^2,2a{t_1}} \right),\left( {at_2^2,2a{t_2}}
\right)$and the focus is$\left( {a,0} \right)$
Three points are collinear, so slopes will be same,
$
\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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Name the Largest and the Smallest Cell in the Human Body ?

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

A sample of an ideal gas is expanded from 1dm3 to 3dm3 class 11 chemistry CBSE
