Answer:
[tex]16x^2+25y^2-400=0[/tex]
Step-by-step explanation:
Given:
- The sum of the distance of any point (x, y) in this conic to the two focuses F₁=(-3,0) and F₂=(3, 0) is constant and equal to 10.
There are 4 types of conic sections:
- Circles (one focus).
- Parabolas (one focus).
- Ellipses (two foci).
- Hyperbolas (two foci).
The definition of an ellipse is:
- The set of points in a plane such that the sum of the distance from a point to each focus is constant.
Therefore, the given definition is one for an ellipse.
[tex]\boxed{\begin{minipage}{7.4 cm}\underline{General equation of an ellipse}\\\\$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$\\\\where:\\\phantom{ww}$\bullet$ $(h,k)$ is the center. \\ \phantom{ww}$\bullet$ $(h\pm a,k)$ or $(h,k\pm b)$ are the vertices. \\ \phantom{ww}$\bullet$ $(h\pm c,k)$ or $(h,k\pm c)$ where $c^2=a^2-b^2$.\\\end{minipage}}[/tex]
As the given foci are (-3, 0) and (3, 0), they are on the x-axis. Therefore:
Therefore, the major axis is on the x-axis and the vertices are (h±a, k) and the co-vertices are (h, k±b).
The constant is the major axis which is 2a. Given the constant is 10:
[tex]\implies 2a=10[/tex]
[tex]\implies a=5[/tex]
Since c² = a² - b²:
[tex]\implies 3^2=5^2-b^2[/tex]
[tex]\implies b^2=5^2-3^2[/tex]
[tex]\implies b^2=25-9[/tex]
[tex]\implies b^2=16[/tex]
Therefore, the equation for the conic section (ellipse) is:
[tex]\implies \dfrac{(x-0)^2}{5^2}+\dfrac{(y-0)^2}{16}=1[/tex]
[tex]\implies \dfrac{x^2}{25}+\dfrac{y^2}{16}=1[/tex]
In standard form:
[tex]\implies 25 \cdot \dfrac{x^2}{25}+25 \cdot \dfrac{y^2}{16}=25 \cdot 1[/tex]
[tex]\implies x^2+\dfrac{25}{16}y^2=25[/tex]
[tex]\implies 16 \cdot x^2+16 \cdot \dfrac{25}{16}y^2=16 \cdot 25[/tex]
[tex]\implies 16x^2+25y^2=400[/tex]
[tex]\implies 16x^2+25y^2-400=400-400[/tex]
[tex]\implies 16x^2+25y^2-400=0[/tex]
