Question

NO LINKS!! Please help me with this problem.

What is the standard form of the equation of the ellipse? ​

Answer 1

Answer:

$\dfrac{x^2}{49}+\dfrac{y^2}{36}=1$

Step-by-step explanation:

From inspection of the given graph, we can see that the center of the ellipse is the origin (0, 0) and the longest diameter of the ellipse is across the x-axis.  

Therefore, we can use the general equation of a horizontal ellipse with its center at (0, 0):

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

where:

• center = (0, 0)
• Vertices = (±a, 0)
• Co-vertices = (0, ±b)
• Foci = (±c, 0) where c²=a²−b²
• Major Axis: longest diameter (2a)
• Minor Axis: shortest diameter (2b)
• Major radius: one half of the major axis (a)
• Minor radius: one half of the minor axis (b)

From inspection of the ellipse:

• Vertices = (-7, 0) and (7, 0)
• Co-vertices = (0, 6) and (0, -6)

Therefore, a = 7 and b = 6

Substitute the found values of a and b into the formula:

$\implies \dfrac{x^2}{7^2}+\dfrac{y^2}{6^2}=1$

$\implies \dfrac{x^2}{49}+\dfrac{y^2}{36}=1$

Learn more about ellipses here:

https://brainly.com/question/28152346

https://brainly.com/question/28144085

Answer 2

${\qquad\qquad\huge\underline{{\sf Answer}}}$

The given figure shows a Horizontal Ellipse with its centre at origin, and as we observe the figure, we can conclude that :

Length of major axis is :

$\qquad \sf  \dashrightarrow \: 2a = 14$

$\qquad \sf  \dashrightarrow \: a = 7$

and that of minor axis :

$\qquad \sf  \dashrightarrow \: 2b = 12$

$\qquad \sf  \dashrightarrow \: b = 6$

Equation of Ellipse will be ~

$\qquad \sf  \dashrightarrow \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

[ plug in the values ]

$\qquad \sf  \dashrightarrow \: \dfrac{ {x}^{2} }{ {7}^{2} } + \dfrac{ {y}^{2} }{ {6}^{2} } = 1$

$\qquad \sf  \dashrightarrow \: \dfrac{ {x}^{2} }{ {49}^{} } + \dfrac{ {y}^{2} }{ {36}^{} } = 1$