Please help answer this thanks

Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{8},0)(± 8 ,0)(, plus minus, square root o

Alchen [17]

Answer:

The equation of ellipse centered at the origin

$\frac{x^2}{18} +\frac{y^2}{10} =1$

Step-by-step explanation:

given the foci of ellipse (±√8,0) and c0-vertices are (0,±√10)

The foci are (-C,0) and (C ,0)

Given data (±√8,0)

the focus has x-coordinates so the focus is lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so c0-vertices are (0,±√10)

given focus C = ae = √8

Given co-vertices ( minor axis) (0,±b) = (0,±√10)

b= √10

The relation between the focus and semi major axes and semi minor axes are $c^2=a^2-b^2$

$a^{2} = c^{2} +b^{2}$

$a^{2} = (\sqrt{8} )^{2} +(\sqrt{10} )^{2}$

$a^{2} =18$

$a=\sqrt{18}$

The equation of ellipse formula

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

we know that $a=\sqrt{18} and b=\sqrt{10}$

Final answer:-

The equation of ellipse centered at the origin

$\frac{x^2}{18} +\frac{y^2}{10} =1$

8 0

3 years ago

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X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square

swat32

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = $-\frac{4}{5}x+\frac{58}{5}$

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

[x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30

(x + 2)² + (y - 5)²- 29 + 20 = 30

(x + 2)² + (y - 5)²- 9 = 30

(x + 2)² + (y - 5)² = 39

By comparing this equation with the standard equation of a circle,

Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

Slope of the line joining this point to the center (-2, 5),

$m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{10-5}{2+2}$

= $\frac{5}{4}$

Let the slope of the tangent which is perpendicular to this line is ' $m_{2}$ '

Then by the property of perpendicular lines,

$m_{1}\times m_{2}=-1$

$\frac{5}{4}\times m_{2}=-1$

$m_{2}=-\frac{4}{5}$

Now the equation of the line passing though (2, 10) having slope $m_{2}=-\frac{4}{5}$

y - y' = $m_{2}(x-x')$

y - 10 = $-\frac{4}{5}(x-2)$

y - 10 = $-\frac{4}{5}x+\frac{8}{5}$

y = $-\frac{4}{5}x+\frac{8}{5}+10$

y = $-\frac{4}{5}x+\frac{58}{5}$

Therefore, equation of the line will be, y = $-\frac{4}{5}x+\frac{58}{5}$

7 0

3 years ago

Find the perimeter and the area of the given figure: PLSSSSSSSSSSSSSSSSSS

Lera25 [3.4K]

Answer:

perimeter: 26

area: 16 + 10 =26

perimeter: add all lengths.

area: multiply LxW

5 0

3 years ago

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Fresh Market is selling black truffles for $15 per ounce and white truffles for $35 per ounce. You bought a package that weighed

Katarina [22]

4oz of white truffles, unless you half to use both white and black

6 0

3 years ago

What is 11=m-2 equal when your solving and checking your answer

AfilCa [17]

Add 2 to both sides

11+2 = 13

m=13

4 0

3 years ago