Question

Let a line $y = mx\,\,\,(m > 0)$ intersects the parabola,${y^2} = x$ at a point P, other than origin. Let the tangent to it at point P meets the X-axis at point Q. If the area ((OPQ)=4square units then what is the value of $m$?

Answer 1

Hint: We will solve the line $y = mx\,\,\,(m > 0)$ with the parabola ${y^2} = x$ to get the point of intersection .After getting the point of intersection ,we can find the equation of tangent to parabola using the formula for tangent $ty = x + a{t^2}$.Then we can put the value of $y = 0$to get the coordinates of point Q. Now we know coordinates of point O,P and Q hence ,we can find the area of triangle and put it equal to given value.

Formula Used:
If ${y^2} = 4ax$ is the parabola and P be any point having parameter ‘$t$’ on it, then the equation of tangent to it on that point is given by – $ty = x + a{t^2}$
Any point on parabola ${y^2} = 4ax$can be written in a single variable $t$ as $(a{t^2},2at)$.
Area of triangle ABC having coordinates $({x_1},{y_1})$ , $({x_2},{y_2})$ and $({x_3},{y_3})$ is given as: $\dfrac{1}{2}\left[ {\left( {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right)} \right]$

Complete step by step solution:
Given -line $y = mx\,\,\,(m > 0)$ and parabola ${y^2} = x$. If we compare the given parabola with the standard parabola ${y^2} = 4ax$then
$4a = 1$
$ \Rightarrow a = \dfrac{1}{4}$
We know that any random point on parabola ${y^2} = 4ax$ can be written in single variable $t$ as $(a{t^2},2at)$.
Here $a = \dfrac{1}{4}$
So, any random point can be taken as –
$P \equiv \left( {\dfrac{1}{4}{t^2},2 \times\dfrac{1}{4}t} \right)$
so, $P \equiv \left( {\dfrac{{{t^2}}}{4},\dfrac{t}{2}} \right)$
Now solving the parabola and line –
Put $y = mx$in parabola ${y^2} = x$
${\left( {mx} \right)^2} = x$
On simplifying the equation –
 ${m^2}{x^2} = x$
$ \Rightarrow {m^2}{x^2} - x = 0$
Factoring the terms
$x({m^2}x - 1) = 0$
$ \Rightarrow x = 0,\dfrac{1}{{{m^2}}}$
Case 1: $x = 0$
${y^2} = x$
$ \Rightarrow {y^2} = 0$
$ \Rightarrow y = 0$
∴$(x,y) \equiv (0,0)$
As given P is any point other than origin, hence this case is rejected.
Case 2: $x = \dfrac{1}{{{m^2}}}$
${y^2} = x$
$ \Rightarrow {y^2} = \dfrac{1}{{{m^2}}}$
$ \Rightarrow {y^2} = \dfrac{1}{m}$
∴$(x,y) \equiv (\dfrac{1}{{{m^2}}},\dfrac{1}{m})$
Comparing $(\dfrac{1}{{{m^2}}},\dfrac{1}{m})$ with $\left( {\dfrac{{{t^2}}}{4},\dfrac{t}{2}} \right)$
$\dfrac{{\dfrac{{{t^2}}}{4}}}{{\dfrac{t}{2}}} = \dfrac{{\dfrac{1}{{{m^2}}}}}{{\dfrac{1}{m}}}$
$ \Rightarrow \dfrac{{2{t^2}}}{{4t}} = \dfrac{1}{m}$
$ \Rightarrow \dfrac{t}{2} = \dfrac{1}{m}$
$ \Rightarrow t = \dfrac{2}{m}$ (1)
Now equation of tangent at point P can be calculated using formula -
$ty = x + a{t^2}$
So equation of tangent is -$ty = x + \dfrac{{{t^2}}}{4}$
To get the coordinate of point Q ,
Put $y = 0$
$x = - \dfrac{{{t^2}}}{4}$
Thus point Q is $( - \dfrac{{{t^2}}}{4},0)$.
Now area of $(\Delta OPQ)$ is given by –
$\dfrac{1}{2}\begin{bmatrix}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{bmatrix}$
So, area $((\Delta OPQ))=\dfrac{1}{2}\begin{bmatrix}0&0&1\\{\dfrac{{{t^2}}}{4}}&{\dfrac{t}{2}}&1\\{ - \dfrac{{{t^2}}}{4}}&0&1\end{bmatrix}$
Expanding the determinant along ${R_1}$
area $(\Delta OPQ)=\dfrac{1}{2}(0 + 0 + ( - \dfrac{{{t^3}}}{8})\,)$
area $(\Delta OPQ)= - \dfrac{{{t^3}}}{{16}}$
Taking the modulus
area $(\Delta OPQ)=\dfrac{{{t^3}}}{{16}}$
given area(OPQ)=$4$
so $\dfrac{{{t^3}}}{{16}} = 4$
$ \Rightarrow {t^3} = 64$
$ \Rightarrow t = {64^{\dfrac{1}{3}}}$
$ \Rightarrow t = 4$
But by equation (1) we know that $t = \dfrac{2}{m}$
Thus $\dfrac{2}{m} = 4$
On solving for m
$m = \dfrac{2}{4}$
$ \Rightarrow m = \dfrac{1}{2}$

Hence value of m is $\dfrac{1}{2}$.

Note: In the questions involving line and parabola, it is advised to always start the question using the parametric equations of parabola ,this reduces the number of variables to 1 and equations of tangent and normal are easy to write in that form.