Respuesta :
Answer:
[tex]\frac{(x+9)^2}{144}+\frac{(y-3)^2}{140}=1[/tex]
Step-by-step explanation:
The standard equation of the ellipse is
[tex]\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1\;\cdots(i)[/tex]
where, [tex](\alpha, \beta)[/tex] is the center and a,b are semi axes of the ellipse along x-axis and y-axis respectively.
Given that, [tex](\alpha, \beta)=(-9,3)[/tex] and major axis, [tex]2a=24[/tex].
So, semi major axis, [tex]a=24/2=12[/tex].
Now, focus of the ellipse is (-7,3).
Let [tex]e[/tex] be the eccentricity of the ellipse,
So, [tex]e=\frac c a[/tex]
where [tex]c[/tex] is the distance between the center and focus of the ellipse.
By using the distance formula,
[tex]c=\sqrt{(-9-(-7))^2+(3-3)^2}=2[/tex]
[tex]\Rightarrow e=\frac {2}{12}=\frac{1}{6}\;\cdots(ii)[/tex]
Again, the relationship among [tex]a,b[/tex] and [tex]e[/tex] is
[tex]e=\sqrt{1-\frac{b^2}{a^2}[/tex]
[tex]\Rightarrow \frac{1}{6}=\sqrt{1-\frac{b^2}{(12)^2}[/tex] [from equation (ii)]
[tex]\Rightarrow \frac{1}{36}=\sqrt{1-\frac{b^2}{144}[/tex] [squaring on both the sides]
[tex]\Rightarrow \frac{b^2}{144}=1-\frac{1}{36}[/tex]
[tex]\Rightarrow b^2=\frac{35}{36}\times144=140[/tex]
So, the value of square of semi-minir axis, [tex]b^2=140[/tex].
Hence, from equation (i), the equation of required ellipse is standard form is
[tex]\frac{(x-(-9))^2}{144}+\frac{(y-3)^2}{140}=1[/tex]
[tex]\Rightarrow \frac{(x+9)^2}{144}+\frac{(y-3)^2}{140}=1[/tex]
The equation of the ellipse is [tex]\frac{(x+9)^{2}}{144}+\frac{(y-3)^{2}}{140} = 1[/tex].
The equation of the ellipse in standard form is described below:
[tex]\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1[/tex] (1)
Where:
- [tex]h, k[/tex] - Coordinates of the center.
- [tex]a[/tex] - Major semiaxis length.
- [tex]b[/tex] - Minor semiaxis length.
The length of the minor semiaxis ([tex]b[/tex]) is found by using the following Pythagorean-like relationship:
[tex]b = \sqrt{a^{2}-c^{2}}[/tex] (2)
If we know that [tex](h,k) = (-9,3)[/tex], [tex]a = 12[/tex] and [tex]F_{1}(x,y) = (-7,3)[/tex], then the equation of the ellipse is:
[tex]c = \sqrt{[(-9)-(-7)]^{2}+(3-3)^{2}}[/tex]
[tex]c = 2[/tex]
[tex]b = \sqrt{12^{2}-2^{2}}[/tex]
[tex]b = 2\sqrt{35}[/tex]
Hence, the equation of the ellipse is [tex]\frac{(x+9)^{2}}{144}+\frac{(y-3)^{2}}{140} = 1[/tex].
We kindly invite to check this question on ellipses: https://brainly.com/question/14281133
