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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.

“Complete step-by-step 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:

$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = {\text{ Distance between them }}$ .

“Complete step-by-step 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:

$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = {\text{ Distance between them }}$ .

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