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A high school mathematics problem, elliptic, master to! Ellipse x ^ 2 / A ^ 2, y ^ 2 / b ^ 2 = 1 (a ＞ B ＞ 0), the straight line passing through the left focus f intersects ellipse at two points a and B, the inclination angle of straight line L is 60 °, vector AF = 2, vector FB 1. Calculate the eccentricity of ellipse 2. If AB length is 15 / 4, find the equation of ellipse C
It is known that A.B and P (2.4) are on the parabola y = - 1 / 2x2 + m, and the inclination angles of the lines PA and Pb are complementary. It is proved that the slope of the line AB is a fixed value
Suppose the slope of PA is k, then the slope of Pb is - K, so the linear equation of PA is y-4 = K (X-2). Substituting it into the parabolic equation, we can get x = (- 2k-2) or 2, so the abscissa of a is x = - 2k-2. Similarly, we can get the abscissa of B is x = 2k-2, and then let a (x1, Y1), B (X2, Y2), so Y1 = 6-1 / 2 * X1 * x1, y2 = 6-1 / 2 * x2 * X2, and (y1-y2) / (x1-x2) = - 1 / 2 (x1 + x2) = 2, so the slope of AB is a fixed value of 2
As shown in the figure, in the straight prism abo-a'b'o ', OO ′ = 4, OA = 4, OB = 3, ∠ AOB = 90 °, D is the midpoint of the line segment a'B', P is a point on the side edge BB ', if op ⊥ BD, calculate the angle between op and the bottom AOB (the result is expressed by the value of the inverse trigonometric function)
Let P (3, 0, z), then BD = {& nbsp; − 32 & nbsp;, & nbsp; 2 & nbsp;, & nbsp; 4 & nbsp;}, Op = {& nbsp; 3 & nbsp;, & nbsp; 0 & nbsp;, & nbsp; Z & nbsp; }∵ BD ⊥ OP, ∵ BD ⊥ OP = − 92 + 4Z = 0.z = 98. ∵ BB ′⊥ plane AOB, ∵ POB is the angle between OP and bottom AOB. Tan ∵ POB = 38, ∵ POB = arctan 38
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