Find the equation in standard form of the ellipse with foci (0, 5) and (0, -5) and major axis of length 14.
a) (x²/24)+(y²/49)=1 = 1
b) (x²/49)+(y²/24)=1 = 1
c) (x²/4)+(y²/49)=1 = 1
d) (x²/49)+(y²/4)=1 = 1
Find the equation in standard form of the ellipse with center at (4, 2), foci (1, 2) and (7, 2), and minor axis of length 10.
a) ((x-4)²/34)+((y-2)²/25)=1 = 1
b) ((x+4)²/25)+((y+2)²/34)=1 = 1
c) ((x-4)²/25)+((y-2)²/34)=1 = 1
d) ((x+4)²/34)+((y+2)²/25)=1 = 1
Please and Thank You !
a) (x²/24)+(y²/49)=1 = 1
b) (x²/49)+(y²/24)=1 = 1
c) (x²/4)+(y²/49)=1 = 1
d) (x²/49)+(y²/4)=1 = 1
Find the equation in standard form of the ellipse with center at (4, 2), foci (1, 2) and (7, 2), and minor axis of length 10.
a) ((x-4)²/34)+((y-2)²/25)=1 = 1
b) ((x+4)²/25)+((y+2)²/34)=1 = 1
c) ((x-4)²/25)+((y-2)²/34)=1 = 1
d) ((x+4)²/34)+((y+2)²/25)=1 = 1
Please and Thank You !
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c = 5 and a = 7
a^2 = b^2 + c^2
49 = b^2 + 25
b^2 = 24
b
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c = 3 and b = 5
a^2 = 25 + 9
a^2 = 34
a
Why do you have =1 written twice?
a^2 = b^2 + c^2
49 = b^2 + 25
b^2 = 24
b
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c = 3 and b = 5
a^2 = 25 + 9
a^2 = 34
a
Why do you have =1 written twice?
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A picture helps.
Draw in the foci and major axis along the y-axis. Place a point on the x-axis representing one of the as yet unknown x-intercepts of the ellipse.
The distance to a focus, 5 in problem 1, and the unknown distance x are two legs of a right triangle. The hypotenuse is half the major axis, 7 in problem 1. Use the Pythagorean formula to solve for x.
x² + y² = 7²
x² + 5² = 7²
x² + 25 = 49
x² = 24
Choose the appropriate answer.
For number 2, draw a picture.
Use the same right triangle analysis to come up with the distance from the center to the vertex.
Half the minor axis (5) is one leg of the right triangle and the distance from the center to a focus (3) is the other.
x² + y² = r²
3² + 5² = r²
9 + 25 = r²
34 = r²
(x - 4)²/34 + (y - 2)²/25 = 1
Draw in the foci and major axis along the y-axis. Place a point on the x-axis representing one of the as yet unknown x-intercepts of the ellipse.
The distance to a focus, 5 in problem 1, and the unknown distance x are two legs of a right triangle. The hypotenuse is half the major axis, 7 in problem 1. Use the Pythagorean formula to solve for x.
x² + y² = 7²
x² + 5² = 7²
x² + 25 = 49
x² = 24
Choose the appropriate answer.
For number 2, draw a picture.
Use the same right triangle analysis to come up with the distance from the center to the vertex.
Half the minor axis (5) is one leg of the right triangle and the distance from the center to a focus (3) is the other.
x² + y² = r²
3² + 5² = r²
9 + 25 = r²
34 = r²
(x - 4)²/34 + (y - 2)²/25 = 1
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OK. Since the major axis appears to be the y-axis in #1 I'm going with a as my answer...a vertical ellipse.
In number 2 the equation has to be either a or c and this is a horizontal ellipse...so I'm going with a as my answer.
I also don't know why you have the =1 =1 in your question.
In number 2 the equation has to be either a or c and this is a horizontal ellipse...so I'm going with a as my answer.
I also don't know why you have the =1 =1 in your question.