If the center of the hyperbola is known to be at the coordinate origin, the focus is on the x-axis, and an asymptote is a straight line 3 / 4x-y = 0, then the eccentricity of the hyperbola is equal to?

If the center of the hyperbola is known to be at the coordinate origin, the focus is on the x-axis, and an asymptote is a straight line 3 / 4x-y = 0, then the eccentricity of the hyperbola is equal to?

If the focus is on the x-axis, then the asymptote equation of hyperbola should be y = ± (B / a) X;
So the centrifugal rate e = C / a = √ (A & # 178; + B & # 178;) / a = √ [(1 + (B / a) &# 178;] = √ [(1 + (3 / 4) &# 178;] = 5 / 4 = 1.25;
Given that the ellipse x2 / A2 + Y2 / B2 = 1, the chord length passing through the focus perpendicular to the major axis is 1, and the focus and the two ends of the minor axis form an equilateral triangle, find the elliptic equation; the line L passing through the point Q (- 1,0) intersects the ellipse with two points a and B, the intersection x = 4 and the point E, the proportion of the q-component vector ab of the point is λ, and the proportion of the e-Component vector ab of the point is μ, find the value of λ + μ
In the first question, y = (b ^ 2) / A is obtained by taking the abscissa of focus C into the equation, that is, the chord length of focus 1 = 2 (b ^ 2) / A is simplified, that is, a = 2B ^ 2
The point from the focus to the vertex of the minor axis is a, because it is an equilateral triangle, so a = 2B synthesize the above two solutions to get b = 1, a = 2 elliptic equation
In the second question, with undetermined coefficients, we can deduce that if the slope of the straight line is k, the linear equation y = K (x + 1) is combined with the ellipse, and then we can write out the fractional ratio formula. The steps are not convenient for calculation, so they are omitted
If the ellipse x234 + y2n2 = 1 and hyperbola x2n2 − y216 = 1 have the same focus, then the value of real number n is ()
A. ±5B. ±3C. 5D. 9
Ellipse x234 + y2n2 = 1, then ﹣ C1 = 34 − n & nbsp; 2, ﹣ focal coordinate is (34 − n & nbsp; 2, 0) (- 34 − n & nbsp; 2, 0), hyperbola: x2n2 − y216 = 1, then half focal length C2 = n & nbsp; 2 + 16 ﹣ 34 − n & nbsp; 2 = n & nbsp; 2 + 16, then real number n = ± 3, so choose B
On the equation AX & # 178; - (2a + 1) x + (a + 1) = 0
On the equation AX & # 178; - (2a + 1) x + (a + 1) = 0
There is
[ax-(a+1)](x-1)=0
therefore
X = 1 or x = (a + 1) / A
(a≠0)
If a = 0, the equation becomes:
-x+1=0
X=1
Square root problem
√5/3 ÷√1/3
Is it 5 / 3 under the root divided by 1 / 3 under the root?
That is 5 / 3 under the root sign divided by 1 / 3 = 5 / 3 under the root sign * 3 = √ 5
2. 2 3 6
To be exact, it's root 5
It is proved that the equation (a2-8a + 20) x2 + 2aX + 1 = 0 is a quadratic equation with one variable no matter what value a takes
It is proved that: ∵ a2-8a + 20 = (A-4) 2 + 4 ≥ 4, ∵ no matter what value a takes, a2-8a + 20 ≥ 4, that is, no matter what value a takes, the quadratic coefficient of the original equation will not be equal to 0, ∵ the equation about X (a2-8a + 20) x2 + 2aX + 1 = 0, no matter what value a takes, the equation is quadratic equation of one variable
On the equation AX & # 178; + 2x + 2A = 0 of X
(1) If one of the equations is larger than 1 and the other is smaller than 1, the range of a is obtained
(2) If the roots of the equation are all greater than 1, the range of a is obtained
First of all, the discriminant = 4-8a ^ 2 > = 0, so - √ 2 / 20, ---- ③
The range of a is - √ 2 / 2
The calculation of square root
X ^ 2 = 49 calculation process
x²＝49
x＝±7
You can shift the positive and negative terms (x = 7-7, x = 7-7)
Let m and n be two real roots of the equation x2-2ax + A + 6 = 0, then the minimum value of (m-1) 2 * (n-1) 2 is
urgent
Eight
According to Weida's theorem
x1+x2=2a
x1x2=a+6
(x1-1)^2+(x2-1)^2
=x1^2-2x1+1+x2^2-2x2+1
=(x1+x2)^2-2x1x2-2(x1+x2)+2
=4a^2-2(a+6)-2*2a+2
=4a^2-6a-10
=4(a-3/4)^2-12.25
Discriminant = 4A ^ 2-4 (a + 6) = 4A ^ 2-4a-24 > = 0
a^2-a-6>=0
(a-3)(a+2)>=0
a> = 3 or a
Solve the equation about X: ax & # 178; + (2a-1) x + a = 0
Ax & # 178; + (2a-1) x + a = 0case 1: if a = 0x = 0case 2: if a is not equal to 0x = {- (2a-1) + √ [(2a-1) ^ 2-4a ^ 2]} / (2a) or {- (2a-1) - √ [(2a-1) ^ 2-4a ^ 2]} / (2a) = {- (2a-1) + √ (- 4A + 1)} / (2a) or {- (2a-1) - √ (- 4A + 1)} / (2a)