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The center is at the origin, the focus is on the y-axis, and the focal length is (3,0) Such as the title
Let's first set up the elliptic equation with focus on the y-axis
Because the focal length is (3,0), C is equal to 3
That is, A-B equals 9
Then bring the point that the ellipse passes through to the y-axis
The elliptic equation on can get a relation between a and B
If we combine it with A-B equal to 9, we can get a
And the value of B, and then into the set focus on the y-axis ellipse
Equation, you can solve the equation
It's very simple. This is my analysis and solution,
Understand
Your question is incomplete, so you can only say so much
Find the equation of the ellipse which passes through the point (2, - 3) and has the same focus as the ellipse 9x ^ 2 + 4Y ^ 2 = 36
x^2/4+y^2/9=1
Intersection coordinates (0, ± root 5)
Let x ^ 2 / A + y ^ 2 / (a + 5) = 1, a > 0
Because through point (2, - 3)
4/a+9/(a+5)=1
4a+20+9a=a^2+5a
a^2-8a-20=0
(a-10)(a+2)=0
a=10
So x ^ 2 / 10 + y ^ 2 / 15 = 1
The ellipse standard equation with focal length of 8 and the sum of distances from any point P to two focal points of the ellipse of 10
2c=8
2a=10
therefore
a^2=25
c^2=16
therefore
b^2=a^2-c^2=9
So the equation is
X ^ 2 / 25 + y ^ 2 / 9 = 1 or x ^ 2 / 9 + y ^ 2 / 25 = 1
It is known that F1 (- C, 0) and F2 (C, 0) are the left and right focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1. Making a straight line L with an inclination angle of 60 degrees through F1 is called the ellipse at two points a and B
The radius of the inscribed circle of the triangle ABF 2 is 2 / 7 times C of the root sign 3. Find (1) the eccentricity of the ellipse; (2) if labl = 8 times root sign 2, find the standard equation of the ellipse
I can't ask the first question. I'm very frustrated,
First, according to the equal area, RL / 2 = 2A. | y1-y2 | / 2. Here R is the radius of the inscribed circle, l is the circumference of the triangle abf2, which is equal to 4a. Then, let AF1 = L1, BF1 = L2, let a be above the x-axis. Then lcos60 ° + - C -- C Λ 2 / a Λ 2 = L / E. we get L1 = 2B Λ 2 / a (2-e). Similarly, L2 = 2B Λ 2
RELATED INFORMATIONS
- 1. The left and right focal points of the ellipse x2a2 + y2b2 = 1 (a > b > 0) are F1 and F2 respectively. The intersection of the straight line with an inclination angle of 120 ° through F2 and the ellipse is m. if MF1 is perpendicular to the X axis, the eccentricity of the ellipse is () A. 12−2311B. 2-3C. 2（2-3）D. 33
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