Question

Need this question now I think it's only a good answer

Mrsus asked Aug 25, 2023

39,664 views

1 Answer

Gel · Answer 1 · 2023-09-01T08:16:12+0000

✒️EQUATION

For ellipses:

-First, let's use completing the square method to determine values of a(semi-major axis), b(semi-minor axis), and the coordinates of the center of the ellipse.

$\mathsf{x^2+4y^2-10x-24y+45=0}$

$\mathsf{x^2-10x+4y^2-24y=-45}$

$\mathsf{(x^2-10x)+4(y^2-6y)=-45}$

$\mathsf{(x^2-10x+25)+4(y^2-6y+9)=-45+25+4(9)}$

$\mathsf{(x-5)^2+4(y-3)^2=16}$

$\mathsf{((x-5)^2)/(16)+((y-3)^2)/(4)=1}$

$\mathsf{((x-5)^2)/((4)^2)+((y-3)^2)/((2)^2)=1}$

-The center of the ellipse is at C(5, 3), semi-major axis, a = 4, and semi-minor axis, b = 2.

-Since the major axis of the ellipse is horizontal, the distance between the center of the ellipse and the co-vertices of the ellipse is ±b.

$\mathsf{CoV_1=(5,3+b)}$

$\mathsf{CoV_2=(5,3-b)}$

-The coordinates of the co-vertices of the ellipse are:

$\mathsf{CoV_1=(5,3+2)} \longrightarrow \mathsf{CoV_1=(5,5)}$

$\mathsf{CoV_2=(5,3-2)} \longrightarrow \mathsf{CoV_2=(5,1)}$

For the parabola:

-We have the following data:

The parabola is opening to the left and axis symmetry along the major axis of the ellipse (axis of symmetry: y = 3):

$\mathsf{(y-k)^2=-4a(x-h)}$

The focus of the parabola is the center of the ellipses.

F = (5, 3)

-The parabola is passing through the co-vertices of the ellipse:

-The parabola is passing through poînts (5, 5) and (5, 1)

-From the first data, we know that the value of k is equal to 3 because the axis of symmetry of the parabola is y = 3.

$\mathsf{(y-3)^2=-4a(x-h)}$

-From the second data, we know that the x-coordinate of the vertex is at h = 5 + a. (Note: the distance of the focus of the parabola and the vertex of the parabola is equal to ±a, since the parabola is opening the the left we use -a)

$\mathsf{F=(5,3)}$

$\mathsf{F=(h-a,k)}$

$\mathsf{5=h-a}$

$\mathsf{h=5+a}$

$\mathsf{V=(h,k)}$

$\mathsf{V=(5+a,3)}$

-Substitute the coordinates of the vertex of the parabola

$\mathsf{(y-3)^2=-4a[x-(5+a)]}$

$\mathsf{(y-3)^2=-4a(x-5-a)}$

$\mathsf{(y-3)^2=-4ax+20a+4a^2}$

-From the third data, we know that if the poînts lies on the parabola, the coordinates of the poînts must satisfy the equation of the parabola.

Using the point (5, 5), x = 5, y = 5:

$\mathsf{(y-3)^2=-4ax+20a+4a^2}$

$\mathsf{(5-3)^2=-4a(5)+20a+4a^2}$

$\mathsf{4=-20a+20a+4a^2}$

$\mathsf{4a^2=4}$

$\mathsf{a^2=1}$

$\mathsf{a=1}$

-Another way to solve for the value of a is using the formula for the length of the latus rectum (LR = 4a). Since the length of the latus rectum is equal to the minor axis of the ellipse, we can easily solve for the value of a.