The left and right vertices of high school mathematics x2 / A2 + Y2 / B2 = 1 (a > b > 0) are A.B. point P is on the ellipse, and is different from two points ab. o is the coordinate origin. (2) if AP = OP, it is proved that the slope k of straight line OP satisfies the absolute value of k > root sign 3

The left and right vertices of high school mathematics x2 / A2 + Y2 / B2 = 1 (a > b > 0) are A.B. point P is on the ellipse, and is different from two points ab. o is the coordinate origin. (2) if AP = OP, it is proved that the slope k of straight line OP satisfies the absolute value of k > root sign 3

It is proved that: AP = OP, P is on the vertical bisector x = - A / 2 of line Ao, and P is on the ellipse
The point P is the intersection of the straight line x = - A / 2 and the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
Take x = - A / 2 into x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 to get y ^ 2 = 3 / 4 * B ^ 2
P (- A / 2, radical 3B / 2) or P (- A / 2, - radical 3B / 2)
Let the slope of the straight line OP be K, then K ^ 2 = (3 / 4 * B ^ 2) / (- A / 2) ^ 2 = 3 (b ^ 2 / A ^ 2)
∵a>b,∴b^2/a^2

Make a vertical line from a point m on the ellipse (a > b > 0) to the x-axis, just pass through the left focus F of the ellipse, and the line AB between the endpoint a of the major axis and the endpoint B of the minor axis is parallel to OM, so as to calculate the eccentricity of the ellipse

(1) ∵ MF1 ⊥ x-axis, ab ∥ OM, ∥ RT △ omf1 ∥ RT △ ABO {mf1bo = of1ao (*) set the point m (- C, Y1), substitute into the elliptic equation x2a2 + y2b2 = 1, get c2a2 + y12b2 = 1, the solution of which is Y1 = B2A (rounding off), so MF1 = B2A, and ∵ Ao = a, Bo = B, of 1 = C, ∵ let Ao, Bo, MF1, of

The ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 is orthogonal to the X axis and a, and a point P and op on the ellipse are perpendicular to AP, so the eccentricity can be calculated

Look

Through any point P on the circle D: x2 + y2 = 4, make two tangents m, n of the ellipse C: x2 / 3 + y2 = 4, and prove m ⊥ n

Is the question wrong
X2 / 3 + y2 = 4,4 should be 1, right?
If the solution is as follows
Let P (x0, Y0) be y-y0 = K (x-x0)
Y = kx-kx0 + Y0, let B = - kX0 + Y0
y=kx+b
The elliptic equation is multiplied by 3 to get X & # 178; + 3Y & # 178; - 3 = 0
Substituting y = KX + B to get
x²+3k²x²+6kbx+3b²-3=0
Delta = 0, delta divided by 4
9k²b²-3b²-9k²b²+9k²+3=0
9k²-3b²+3=0
3k²-b²+1=0
b=-kx0+y0
3k²-k²x0²+2kx0y0-y0²+1
△=3y0²+x0²-3＞0
Two K1 · K2 = (- Y0 & # 178; + 1) / (3-x0 & # 178;)
x0²+y0²=4,y0²=4-x0²
k1·k2=(x0²-3)/(3-x0²)=-1
Two straight lines are vertical at will
m⊥n