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Given the inscribed rectangle ABCD of the ellipse x2 / A2 + Y2 / B2 = 1 (a > b > 0) (ABCD is all on the ellipse), find the maximum face of the rectangle
Obviously, when the area of the inscribed rectangle is the largest, the center coincides with the center of the ellipse, which is the origin. If one of the vertex coordinates is (asinr, bcosr), then the other three vertex coordinates are: (asinr, - bcosr), (- asinr, bcosr), (- asinr, - bcosr). Therefore, the side lengths of the rectangle are: 2asinr, 2bcosr area = 2As
As shown in the figure, f is a focus of the ellipse x2 / A2 + Y2 / B2 = 1 (a > b > 0)
As shown in the figure, f is a focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), a and B are the two vertices of the ellipse, the eccentricity of the ellipse is 1 / 2, the point C is on the x-axis, the circle m determined by BC ⊥ BF, B, C and F is just tangent to the line L1: x + √ 3Y + 3 = 0. (1) the equation of solving the ellipse (2) the intersection of the line L2 passing through point a and the circle m with P and Q, and the product of the vector MP and the vector MQ is negative 2
Ha ha, is the landlord a sophomore? We just finished the exam. I'm bored. I'll practice
I analyzed that B should be the lower vertex and f the left focus
BF slope = - root 3. So KBC = root 3
Because of the coordinates of point B (0, - b), the BC equation can be obtained
According to BC equation and X-axis focus, the coordinate of point C is (root 3B. 0), that is, (3C. 0)
Because f (- c.0). B (- root 3C, 0)
So it is easy to see that the center of circle m is (c.0) and the radius is 2C
Then L1 is tangent to the circle, so the distance from the center of the circle to L1 is equal to the radius
The formula of distance from point to straight line is simplified to C = 1
So a = 2, B = root 3. The elliptic equation is one fourth x 2 + three thirds y 2 = 1
As for the second question, you have to tell me whether point a is the left vertex or the right vertex. Then I will continue to do it
OK. Let's think it's the left focus
Vector MP multiplied by vector MQ = - 2 MP.MQ Is the radius of the circle m, so MP = MQ = 2
Cos angle PMQ = - half. So ∠ PMQ = 120 degrees
So the distance between M and line L2 is rsin30 ° = 1
Let the slope of L2 be K. then y = KX + 2K
The square of K is one eighth
So k = + - 2 times the root one-half
Novice answer. Do not know symbols. Only text. Excuse me
As shown in the figure, a and B are the left and right vertices of the ellipse C: x2 / A2 + Y2 / B2 = 1 (a > b > 0). M is any point on the ellipse different from a and B. if the eccentricity of the ellipse C is 1 / 2, and the equation of the right directrix L is x = 4. Question: let a straight line am intersect l at P, and a circle with a diameter of MP intersects MB at Q. try to prove that the intersection r of a straight line PQ on X axis is a fixed point, and calculate the R coordinate
A ^ 2 / C = 4C / a = 1 / 2A = 2C = 1b ^ 2 = 3x ^ 2 / 4 + y ^ 2 / 3 = 1 let the x-axis of the intersection of the collimators be at t, o be the origin, m coordinate be (XO, yo) at / Pt = (XO + AO) / yopt = atyo / (XO + 2) = 6yo / (XO + 2) P point coordinate be (4,6yo / (XO + 2)) MB, the slope is (- yo) / (2-xo) PQ is perpendicular to MB, so PQ slope is (2-xo) / yo, and MB is oblique
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