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If the eccentricities passing through hyperbola x2 / a2-y2 / B2 = 1 and ellipse x2 / M2 + Y2 / B2 = 1 (a > 0, M > b > 0) are reciprocal to each other, then the triangle with a, B.M as side length What triangle is it? Can it be an isosceles right triangle?
Is a right triangle
According to the meaning of the title:
Hyperbola: E1 ^ 2 = 1 + B2 / A2
Ellipse: E2 ^ 2 = 1-b2 / m2
e1=-1/e2,∴e1^2*e2^2=1
That is, (1 + B2 / A2) (1-b2 / m2) = 1
The results show that A2 + B2 = m2,
This △ is RT
On the hyperbola x2 / a2-y2 / B2 = 1 (a > 0, b > 0), the absolute value of the distance difference between a point P and its two focal points is 8, and the inclination angle of an asymptote is 0
If the inclination angle is arctan 3 / 4, make a parallel line of one asymptote through p to intersect another asymptote at point m, and calculate the area s of △ OPM
The absolute value of the distance between point P and his two focal points is 8
2a=8
a=4
The inclination angle of asymptote is arctan 3 / 4
The slope of asymptote is 3 / 4
∵x²/a²-y²/b²=1(a>0,b>0)
The asymptote is y = ± B / ax
∴b/a=3/4
∴b=3
The hyperbolic equation is X & # 178 / 16-y & # 178 / 9 = 1
Let P coordinate (x0, Y0)
MP:y=-3/4(x-x0)+y0
Combined with y = 3 / 4x
M（2(x0/4+y0/3),3(x0/4+y0/3)/2)
Distance from P to OM = | 3x0-4y0 | / 5
The area of OPM is s = 1 / 2 * | 3x0-4y0 | / 5 * √ (4 (x0 / 4 + Y0 / 3) &# 178; + 9 (x0 / 4 + Y0 / 3) &# 178 / 4)
=1/2*|3x0-4y0|/5*5/2*|x0/4+y0/3|
=1/4*|( 3x0-4y0)(x0/4+y0/3)|
=1/4*|3x0²/4-4y0²/3|
=1/4*12*|x0²/16-y0²/9|
=3*1
=3
If the distance from the apex of hyperbola C: x2-y2 / B2 = 1 (b > 0) to the asymptote is √ 2 / 2, then the eccentricity of hyperbola e E=
A.2
B.√2
C.3
D.√3
The equation of asymptote is y = ± (B / a) X
In this paper, the right vertex a (1,0) of hyperbola is given. An asymptote is y = BX
Point a (1,0), linear equation bx-y = 0. Now apply [formula of distance from point to line] and you can do it yourself
The distance from the right focus F2 of hyperbola x2 / a2-y2 / B2 = 1 (a > 0, b > 0) to the straight line passing through point a (a, 0), B (0, b) is equal to half the length of the imaginary half axis of hyperbola,
Finding the eccentricity of hyperbola
The equation of line AB is x / A + Y / b = 1, which is reduced to BX + ay AB = 0
|Bc-a * 0-ab | / SQR (a ^ 2 + B ^ 2) = B / 2, then (BC AB) / C = B / 2, C = 2A, e = 2
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