An ellipse has equation 25𝑥^2 + 16𝑦^2 + 150𝑥 − 32𝑦 = 159. Find the standard equations of all parabolas whose vertex is a focus of this ellipse and whose focus is a vertex of this ellipse
First find all the goods on 25x^2 + 16y^2 - 150x - 32y = 159
by completing the square
25(x^2 + 6x + .....) + 16(y^2 - 2y + ....) = 159
25(x^2 + 6x + 9) + 16(y^2 - 2y + 1) = 159 + 225 + 16
25(x+3)^2 + 16(y-1)^2 = 400
divide each term by 400
(x+3)^2 / 16 + (y-1)^2 / 25 = 1
so we have a vertical ellipse, with centre at (-3,1)
vertices at the end of the major axis, those being (-3,5) and (-3,-5)
and at the end of the minor axis at (4,1) and (-4,1)
for the foci:
c^2 = 5^2 - 4^2 = 9
c = ± 3
The two foci would be 3 units above and below the centre(-3,1)
which would be (-3,4) and (-3,-2)
Our parabola must have as a vertex the focus of the ellipse, so one would
be:
y = a(x+3)^2 + 4 , and the other would be y = a(x+3)^2 - 2
the first:
y = a(x+3)^2 + 4
the shape of our parabola is determined by x^2 = 4py ,
(a standard formula) , but we know p = 1
(the difference between the vertex and the focus ) , so x^2 = 4y or y = (1/4)x^2
our parabolas are:
y = (1/4)(x+3)^2 + 4 and y = (1/4)(x+3)^2 - 2
To find the standard equations of all parabolas whose vertex is a focus of the ellipse and whose focus is a vertex of the ellipse, we need to first find the coordinates of the foci and vertices of the ellipse.
The standard form of the equation of an ellipse is given by:
𝑥^2/𝑎^2 + 𝑦^2/𝑏^2 = 1
By comparing this equation with the given equation of the ellipse, we can find the values of 𝑎 and 𝑏.
Thus, we have:
25𝑥^2 + 16𝑦^2 + 150𝑥 − 32𝑦 = 159
To find the values of 𝑎 and 𝑏, we need to rewrite the equation in the standard form of an ellipse:
25𝑥^2 + 150𝑥 + 16𝑦^2 − 32𝑦 = 159
Dividing the equation by 159 to make the right-hand side equal to 1:
(25𝑥^2 + 150𝑥)/159 + (16𝑦^2 − 32𝑦)/159 = 1
Simplifying the equation:
(𝑥^2 + 6𝑥)/6.36 + (𝑦^2 − 2𝑦)/9.94 = 1
Now we can rewrite the equation in terms of 𝑎 and 𝑏:
(𝑥^2 + 6𝑥)/𝑎^2 + (𝑦^2 − 2𝑦)/𝑏^2 = 1
Comparing coefficients, we find:
𝑎^2 = 6.36
𝑏^2 = 9.94
Taking the square root of both sides, we get:
𝑎 = √6.36 ≈ 2.52
𝑏 = √9.94 ≈ 3.15
Now we can find the foci and vertices of the ellipse:
The coordinates of the center of the ellipse are given by (-𝑧, -𝑧), where 𝑧 is a value that makes the right-hand side of the equation equal to 1.
So, the coordinates of the center of the ellipse are (-6/2, -(-2)/2) = (-3, 1).
The distance from the center to each focus is given by 𝑐 = √(𝑎^2 − 𝑏^2).
So, 𝑐 = √(2.52^2 - 3.15^2) ≈ √(6.3504 - 9.9225) ≈ √(-3.5721) which is not a real value.
Since the value of 𝑐 is not real, it means that the given equation does not represent an ellipse. Therefore, there are no parabolas whose vertex is a focus of this supposed ellipse and whose focus is a vertex of this ellipse.