Question

Identify the equations of ellipses whose major axis lengths are twice their minor axis lengths.

Answer 1

Answer:

If its set up like

1 2

3 4

5 6

the answer is 1,2,4,6

Step-by-step explanation:

i hope my math on desmos was right xD

Answer 2

The equations of ellipses$2x^{2} +8y^{2} -12x+16y-174=0$, $4x^{2} +y^{2} +16x+4y+4=0$,$3x^{2}+12y^{2} +18x-24y-69=0$ and $x^{2} +9y^{2} -2x+18y-71=0$ having the major axis lengths are twice their minor axis length.

Equations of ellipse

$\frac{(x-h)^{2} }{a^{2} } +\frac{(y-k)^{2} }{b^{2} } =1$         (for horizontal major axis ,a>b)

$\frac{(x-h)^{2} }{b^{2} } +\frac{(y-k)^{2} }{b^{2} }=1$           ( for vertical major axis, b>a))

Where, (h, k) is the center of ellipse.

What is major axis of an ellipse?

The major axis of an ellipse is its longest diameter(a line that runs through the center and both foci, with ends at the two most widely separated points  on perimeter).

Formula for calculating the length of major axis

length of major axis = 2a       (where a is the length of semi major axis)

What is minor axis of an ellipse?

The minor axis of an ellipse is the line that contains the shorter of the two line segments about which the ellipse is symmetrical.

Formula for calculating the length of minor axis

length of minor axis = 2b               (where b is the length of semi minor axis)

According to the given question

we have, the equations  of ellipses

A) $(4x^{2} +32x)+(25y^{2} -250y)+589=0$

$4(x^{2} +8x)+25(y^{2} -10y)+589=0$

$4(x^{2} +8x+b^{2} )-4b^{2} +25(y^{2} -10y+b^{2} )-25b^{2} +589=0$

$4(x^{2} +8x+4^{2})-4(4^{2})+25(y^{2} -10y+b^{2} )-25(5^{2} ) +589=0$

$4(x+4)^{2}+25((y-5)^{2} )-100=0$

$4(x+4)^{2} +25(y-5)^{2} =100$

dividing both the sides by 100

$4\frac{(x+4)^{2} }{100} +25\frac{(y-5)^{2} }{100} =1$

$\frac{(x+4)^{2} }{25} +\frac{(y-5)^{2} }{4} =1$

$\frac{(x+4)^{2} }{5^{2} } +\frac{((y-5)^{2} }{2^{2} }=1$

here, h=-4 and k=5

major axis =2a=2×5=10

minor axis=2b=2×2=4

⇒ the length of major axis is not twice the length of minor axis.

Similarly, we will do for the other equations

B).

$2x^{2} +8y^{2} -12x+16y-174=0\\\frac{(x-3)^{2} }{10^{2} } +\frac{(y+1)^{2} }{5^{2} } =1$

here, h=3 and k=-1

major axis:2a=2×10=20

minor axis:2b=2×5=10

⇒ length of major axis is twice the length of minor axis.

C).

$4x^{2} +y^{2} +16x+4y+4=0$

$\frac{(x+2)^{2} }{2^{2} } +\frac{(y+2)^{2} }{4^{2} } =1$

here, h=-2 and k= -2

major axis: 2b=2×4=8

minor axis: 2a= 2×2=4

⇒ the length of the major axis is twice of the length of minor axis.

D).

$x^{2} +4y^{2} +6x-8y-23=0$

$\frac{(x+3)^{2} }{6^{2} } +\frac{(y-1)^{2} }{3^{2} } =1$

here, h=-3 and k=1

major axis: 2a=2×6=12

minor axis: 2b= 2×3=6

⇒ the length of major axis is twice of the length of minor axis.

E).

$16x^{2} +y^{2} -64x+8y+16=0$

⇒$\frac{(x-2)^{2} }{2^{2} } +\frac{(y+4)^{2} }{2^{2}} =1$

here, h=2 and k=-4

major axis: 2a=  4

minor axis :2b=4

⇒ the length of major axis is not the twice of minor axis.

F)

$x^{2} +9y^{2} -2x+18y-71=0$

⇒$\frac{(x-1)^{2} }{9^{2} } + \frac{(y+1)^{2} }{3^{2} } =1$

here, h=1,  and k=-1

major axis: 2a= 18

minor axis: 2b = 9

⇒ the length of major axis is twice to the length of minor axis.

Learn more about the major and minor axis of the ellipse here:

https://brainly.com/question/14180045

#SPJ2