Given that the ellipse x2a2 + Y29 = 1 (a > 0) and hyperbola x24 − Y23 = 1 have the same focus, then the value of a is () A. 2B. 10C. 4D. 10

Given that the ellipse x2a2 + Y29 = 1 (a > 0) and hyperbola x24 − Y23 = 1 have the same focus, then the value of a is () A. 2B. 10C. 4D. 10

Hyperbolic equation is changed to x24 − Y23 = 1, (1 point) thus a = 2, B = 3, (3 points) C = 7, the focus is (- 7, 0), (7, 0). (7 points) in the ellipse, then A2 = B2 + C2 = 9 + 7 = 16. (11 points) then the value of a is 4

Ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 a > b > o the right focus is F. the intersection of its right guide line and X axis is a. there is a point P on the ellipse which satisfies the vertical bisection of line AP
The ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 a > b > o the right focus is F. the intersection of the right guide line and the X axis is a. there is a point P on the ellipse. If the vertical bisector of the line AP passes through F, then the value range of E is?

From the known | PF | = | AF | = a ^ / C - C = B ^ 2 / C, let P (x0, Y0) then - a ≤ x0 ≤ a... ① make pH vertical right collimator to h through P, then | pH | = a ^ 2 / C - x0, according to the definition of ellipse eccentricity e = | PF | / | pH | = (b ^ 2 / C) / (a ^ 2 / C - x0), it is concluded that: a (AC-B ^ 2) / C ^ 2 = x0 from the definition of - a ≤ a (AC-B ^ 2) / C ^ 2 ≤ a

What is the minimum value of the sum of the distances from a point m to the fixed points c (3,1) and B (2,0) on the hyperbola x ^ 2-y ^ 2 = 2?

We can see that point B is the focus of hyperbola, so the distance from m to B is equal to the distance d from m to the Quasilinear x = 1
So the sum of the distances from m to C (3,1) and B (2,0) is equal to D plus the distance from m to C
From the vertical line at point C to the guide line x = 1, let the intersection point be point D, and the distance d = 3-1 = 2, we can see that it is the shortest distance (the sum of the two sides of the triangle is greater than the third side)
So the minimum is 2