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If an isosceles right triangle is formed by one vertex and two focal points of an ellipse, what is the eccentricity of the ellipse
One vertex and two focuses of the ellipse form an isosceles right triangle, that is, B = C
That is, a = √ (B & # 178; + C & # 178;) = (√ 2) C
Eccentricity = C / a = (√ 2) / 2
Given that the image of inverse scale function y = 1-2m / X passes through point a (a, - 2A), then the value range of M is
That is - 2A = (1-2m) / A
1-2m=-2a²
a≠0
So a & # 178; > 0
-2a
If we know the point in the fourth quadrant of the intersection of the images of two linear functions y = 2 (x-a) - 4 and y = - 3 (x + 1) + 4a, what is the value range of a?
Please help me with the detailed process. Thank you
y=2x-2a-4=-3x-3+4a
x=(6a+1)/5
In the fourth quadrant, x > 0, Y0
a>-1/6
y=2x-2a-4=2(6a+1)/5-2a-4
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