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Question: The focal distance of a point on the parabola is \({y^2} = 8x{\text{ is 4}}\); Find the coordinates ...

The focal distance of a point on the parabola is y2=8x is 4{y^2} = 8x{\text{ is 4}}; Find the coordinates of the point?

Explanation

Solution

Hint – First of all, read the question carefully and write the things given in the question i.e. focal distance of the parabola is 44 i.e. D = 4{\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 P(x,y){\text{P}}\left( {x,y} \right). The equation of parabola i.e. y2=8x{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. y2=4ax{y^2} = 4ax to solve this given problem.
So, compare the given equation y2=8x{y^2} = 8x with the standard equation of parabola i.e. y2=4ax{y^2} = 4ax, then we will find that a = 2{\text{a = 2}} by comparing 8x and 4ax8x{\text{ and }}4ax.
As we know that the standard focus of the parabola is (a,0)\left( {a,0} \right). Hence, the focus F{\text{F}} of the given parabola is (2,0)\left( {2,0} \right).
According to the question D = 4{\text{D = 4}} and we assumed the point on the locus as P(x,y){\text{P}}\left( {x,y} \right).
By using the distance the formula on F(2,0) and P(x,y){\text{F}}\left( {2,0} \right){\text{ and P}}\left( {x,y} \right) we will get ,
4 = (x−2)2+(y−0)2{\text{4 = }}\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 0} \right)}^2}}
By squaring on both sides,
16=(x−2)2+(y)216 = {\left( {x - 2} \right)^2} + {\left( y \right)^2}
Now, put the value of y2{y^2} as 8x8x which is given in question and expand the equation (x−2)2{\left( {x - 2} \right)^2}
16=x2+4−4x+8x16 = {x^2} + 4 - 4x + 8x
x2+4x−12=0{x^2} + 4x - 12 = 0
By using factorisation method,
x2+6x−2x−12=0{x^2} + 6x - 2x - 12 = 0
x(x+6)−2(x+6)=0x\left( {x + 6} \right) - 2\left( {x + 6} \right) = 0
(x−2)(x+6)=0\left( {x - 2} \right)\left( {x + 6} \right) = 0
This implies that xx can be 2,−62, - 6 but according to the equation y2=4ax{y^2} = 4ax, xx cannot be negative.
So, we left with only x=2x = 2
Now, by putting the value of xx in y2=8x{y^2} = 8x
We will get,
y2=8×2{y^2} = 8 \times 2
y2=16{y^2} = 16
Apply square root both sides, we will get
y=4,−4y = 4, - 4
So, the coordinates are (2,4) and (2,−4)\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:
(1) y2 = 4ax  (2) y2=−4ax (3) x2=4ay (4) x2=−4ay  \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. (x1,y1) and (x2,y2)\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right) is:
(x2−x1)2+(y2−y1)2= Distance between them \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = {\text{ Distance between them }} .