Question

What are the center and vertices of the ellipse given by (x-10)^2/49+(y+4)^2/4=1?

Answer 1

The center of the ellipse is (10 , -4) and its vertices are (17 , -4) , (3 , -4)

Step-by-step explanation:

The standard form of the equation of an ellipse with center (h , k)  

and major axis parallel to x-axis is $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$  

• The coordinates of the vertices are (h ± a , k )  
• The coordinates of the foci are (h ± c , k), where c² = a² - b²  

∵ The equation of the ellipse is $\frac{(x-10)^{2}}{49}+\frac{(y+4)^{2}}{4}=1$

- Compare it by the standard form of the equation of an ellipse

∴ h = 10 and k = -4

∴ a² = 49 and b² = 4

∵ The center of the ellipse is (h , k)

∴ The center of the ellipse is (10 , -4)

∵ Its vertices are (h + a , k) and (h - a , k)

∵ a² = 49

- Take square root for both sides

∴ a = 7

∵ h = 10 and k = -4

- Substitute the values of h , a , k in the vertices above

∴ Its vertices are (10 + 7 , -4) and (10 - 7 , -4)

∴ Its vertices are (17 , -4) and (3 , -4)

The center of the ellipse is (10 , -4) and its vertices are (17 , -4) ,

(3 , -4)

Learn more:

You can learn more about the conics in brainly.com/question/4054269

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