Question

The line \[x - b + \lambda y = 0\] cuts a parabola \[{y^2} = 4ax\]at \[P(a{t_1}^2,2a{t_1})\] and\[Q(a{t_2}^2,2a{t_2})\]. If \[b \in [2a,4a]\] and\[\lambda \in \Re \], then \[{t_1}{t_2}\] belongs to
A) \[[ - 4, - 2]\]
B) \[[ - 4, - 3]\]
C) \[[ - 3, - 2]\]
D) None of these

Answer 1

Hint: Here we are going to consider a point in the parabola and substitute it in the given line and form a quadratic equation from it and using the relation between roots and coefficient we will find the required range.

Formula used:
Any point on a parabola of equation \[{y^2} = 4ax\]can be taken as\[(a{t^2},2at)\].
Let us consider, \[\alpha ,\beta \] be two roots of a quadratic equation\[a{x^2} + bx + c = 0\], then from the relation between roots and coefficient we get, \[\alpha + \beta = \dfrac{{ - b}}{a}\] and\[\alpha \beta = \dfrac{c}{a}\].

Complete step by step solution:
The diagrammatic representation of given equations is,

It is given that, the line \[x - b + \lambda y = 0\] cuts a parabola \[{y^2} = 4ax\]at \[P(a{t_1}^2,2a{t_1})\] and\[Q(a{t_2}^2,2a{t_2})\].
It is given also that, \[b \in [2a,4a]\] and\[\lambda \in \Re \].
Now, we have to find out the range of\[{t_1}{t_2}\].
We know that, the any point on a parabola \[{y^2} = 4ax\]can be taken as \[(a{t^2},2at)\]
Since, the line \[x - b + \lambda y = 0\] cuts a parabola \[{y^2} = 4ax\]at \[(a{t^2},2at)\]
Let us substitute \[x = a{t^2}y = 2at\] in the equation of the line we get,
\[a{t^2} - b + \lambda (2at) = 0\]
Let us rearrange the expression and mark it as equation (1) we get,
\[a{t^2} + \lambda (2at) - b = 0\]… (1)
Since, the line \[x - b + \lambda y = 0\] cuts a parabola \[{y^2} = 4ax\]at \[P(a{t_1}^2,2a{t_1})\] and\[Q(a{t_2}^2,2a{t_2})\], \[{t_1}\& {t_2}\] be the roots of the above equation (1).
Hence, by the relation between roots and coefficient given in the hint we get,
\[{t_1}{t_2} = \dfrac{{ - b}}{a}\]
Also it is given that, the range of \[b\] is \[[2a,4a]\] which means that $b$ lies between $2a$ and $4a$,
That is \[2a \le b \le 4a\]
Let us divide the above inequality by \[a\] we get,
\[2 \le \dfrac{b}{a} \le 4\]
Also let us multiply \[ - 1\] with each element in the inequality we get,
\[ - 4 \le \dfrac{{ - b}}{a} \le - 2\]
We initially have that, \[{t_1}{t_2} = \dfrac{{ - b}}{a}\]
So the above inequality becomes \[ - 4 \le {t_1}{t_2} \le - 2\]
So, the range of \[{t_1}{t_2}\] is \[[ - 4, - 2]\]
Hence \[{t_1}{t_2} \in [ - 4, - 2]\].

$\therefore$The correct option is (A) \[[ - 4, - 2]\]

Note:
Let us consider, \[x\] be a real number in the range\[[a,b]\], so the range of \[ - x\] is\[[ - b, - a]\].
If the negative sign will be added, the range will be interchanged.
In other words, the inequality changes when it is multiplied by -1, that is if inequality is multiplied by -1 if there is greater than then it is changed to less than and vice versa.