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# The focal distance of a point on the parabola is ${y^2} = 8x{\text{ is 4}}$; Find the coordinates of the point?

Last updated date: 15th Jul 2024
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Hint – First of all, read the question carefully and write the things given in the question i.e. focal distance of the parabola is $4$ i.e. ${\text{D = 4}}$ which is the distance between the focus and that particular point on the parabola and let the given point and coordinates be ${\text{P}}\left( {x,y} \right)$. The equation of parabola i.e. ${y^2} = 8x$. Now, this will give us a clear picture to understand the question. Thus we will get our desired answer.

Now, we will find the coordinates of that particular point. We will use the standard parabola equation i.e. ${y^2} = 4ax$ to solve this given problem.
So, compare the given equation ${y^2} = 8x$ with the standard equation of parabola i.e. ${y^2} = 4ax$, then we will find that ${\text{a = 2}}$ by comparing $8x{\text{ and }}4ax$.
As we know that the standard focus of the parabola is $\left( {a,0} \right)$. Hence, the focus ${\text{F}}$ of the given parabola is $\left( {2,0} \right)$.
According to the question ${\text{D = 4}}$ and we assumed the point on the locus as ${\text{P}}\left( {x,y} \right)$.
By using the distance the formula on ${\text{F}}\left( {2,0} \right){\text{ and P}}\left( {x,y} \right)$ we will get ,
${\text{4 = }}\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 0} \right)}^2}}$
By squaring on both sides,
$16 = {\left( {x - 2} \right)^2} + {\left( y \right)^2}$
Now, put the value of ${y^2}$ as $8x$ which is given in question and expand the equation ${\left( {x - 2} \right)^2}$
$16 = {x^2} + 4 - 4x + 8x$
${x^2} + 4x - 12 = 0$
By using factorisation method,
${x^2} + 6x - 2x - 12 = 0$
$x\left( {x + 6} \right) - 2\left( {x + 6} \right) = 0$
$\left( {x - 2} \right)\left( {x + 6} \right) = 0$
This implies that $x$ can be $2, - 6$ but according to the equation ${y^2} = 4ax$, $x$ cannot be negative.
So, we left with only $x = 2$
Now, by putting the value of $x$ in ${y^2} = 8x$
We will get,
${y^2} = 8 \times 2$
${y^2} = 16$
Apply square root both sides, we will get
$y = 4, - 4$
So, the coordinates are $\left( {2,4} \right){\text{ and }}\left( {2, - 4} \right)$

Note – In this type of questions, firstly we should compare the given equation with the standard parabolic equations which are:
$\left( 1 \right){\text{ }}{y^2}{\text{ = 4}}ax{\text{ }} \\ \left( 2 \right){\text{ }}{{\text{y}}^2} = - 4ax \\ \left( 3 \right){\text{ }}{x^2} = 4ay \\ \left( 4 \right){\text{ }}{x^2} = - 4ay \\$
Then simply putting those values in the equation we get our required answer.
Do note that the distance formula between the two points i.e. $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ is:
$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = {\text{ Distance between them }}$ .