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Question: How do you write the equation given focus \(( - 4, - 2)\) and directrix \(x = - 8\)?...

How do you write the equation given focus (−4,−2)( - 4, - 2) and directrix x=−8x = - 8?

Explanation

Solution

The relation between the focus and directrix is the distance between the focus and the point is equal to the directrix equation. The points are unknown and by solving the equalities we can find the equation with two unknowns x, y. The distance between the two points can be calculated by taking the square root of the sum of squares of difference between the axis points.

Formula:
The focus points and another point are given,
Focus points are (a,b)(a,b).
points are (x,y)(x,y).
d=(a−x)2+(b−y)2d = \sqrt {{{(a - x)}^2} + {{(b - y)}^2}}
While directrix is given
x=x0x = {x_0}
d=x0+ad = {x_0} + a
(a+b)2=a2+2ab+b2{(a + b)^2} = {a^2} + 2ab + {b^2}

Complete step-by-step answer:
Given,
The focus points given are (−4,−2)( - 4, - 2)and the directrix equation is x=−8x = - 8.
We need to find the equation.
To find the equation with two unknown such as x and y, we have to find the distance between the focus and the unknown point.
Given values are focus (−4,−2)( - 4, - 2).
Let us assume the points as (x0,y0)({x_0},{y_0}).
We know the formula to find the distance.
Focus points are (a,b)(a,b).
points are (x,y)(x,y).
d=(a−x)2+(b−y)2d = \sqrt {{{(a - x)}^2} + {{(b - y)}^2}}
By comparing the given values,
a=−4a = - 4, b=−2b = - 2, x=x0x = {x_0} and y=y0y = {y_0}
By substituting in the equation, we get
d=(−4−x0)2+(−2−y0)2d = \sqrt {{{( - 4 - {x_0})}^2} + {{( - 2 - {y_0})}^2}}
while directrix is given, distance is
d=x0+8d = {x_0} + 8
By equating these, we get
(−4−x0)2+(−2−y0)2=x0+8\sqrt {{{( - 4 - {x_0})}^2} + {{( - 2 - {y_0})}^2}} = {x_0} + 8
By squaring on both sides,
(−4−x0)2+(−2−y0)2=(x0+8)2{( - 4 - {x_0})^2} + {( - 2 - {y_0})^2} = {({x_0} + 8)^2}
By taking common negative sign in terms, we get
(4+x0)2+(2+y0)2=(x0+8)2{(4 + {x_0})^2} + {(2 + {y_0})^2} = {({x_0} + 8)^2}
We know the formula (a+b)2=a2+2ab+b2{(a + b)^2} = {a^2} + 2ab + {b^2}
By expanding the terms in the equation in accordance with the formula, we get
42+2×4×x0+x02+22+2×2×y0+y02=82+2×8×x0+x02{4^2} + 2 \times 4 \times {x_0} + x_0^2 + {2^2} + 2 \times 2 \times {y_0} + y_0^2 = {8^2} + 2 \times 8 \times {x_0} + x_0^2
By simplifying the above terms, we get
16+8x0+x02+4+4y0+y02=64+16x0+x0216 + 8{x_0} + x_0^2 + 4 + 4{y_0} + y_0^2 = 64 + 16{x_0} + x_0^2
By adding the like term, we get,
20+8x0+x02+4y0+y02=64+16x0+x0220 + 8{x_0} + x_0^2 + 4{y_0} + y_0^2 = 64 + 16{x_0} + x_0^2
By bringing the terms in the right side to left side,
20+8x0+x02+4y0+y02−64−16x0−x0220 + 8{x_0} + x_0^2 + 4{y_0} + y_0^2 - 64 - 16{x_0} - x_0^2
By simplifying we get,
4y0+y02−44−8x0=04{y_0} + y_0^2 - 44 - 8{x_0} = 0
By replacing (x0,y0)({x_0},{y_0})with (x,y)(x,y), we get,
4y+y2−44−8x=04y + {y^2} - 44 - 8x = 0
By bringing the x term to the right hand side of the equation
4y+y2−44=8x4y + {y^2} - 44 = 8x
Dividing each term by 88, we get
x=y2+y28−112x = \dfrac{y}{2} + \dfrac{{{y^2}}}{8} - \dfrac{{11}}{2}
By rearranging the above terms, we get
x=y28+y2−112x = \dfrac{{{y^2}}}{8} + \dfrac{y}{2} - \dfrac{{11}}{2}
The equation is x=y28+y2−112x = \dfrac{{{y^2}}}{8} + \dfrac{y}{2} - \dfrac{{11}}{2}.

Note: Always remember that the relation between the focus and directrix is the distance between the focus and the point is equal to the directrix equation. The basic formula must be known correctly. Then always enter the value with the correct sign. Do not forget to replace (x0,y0)({x_0},{y_0})with (x,y)(x,y) because it is more important to have the final answer in (x,y)(x,y).