Point a (x0, Y0) is on the right branch of hyperbola x24 − y232 = 1. If the distance from point a to the right focus is equal to 2x0, then x0=______ ．

Point a (x0, Y0) is on the right branch of hyperbola x24 − y232 = 1. If the distance from point a to the right focus is equal to 2x0, then x0=______ ．

A = 2. C = 6, the right focus f (6, 0) substitutes a (x0, Y0) into the hyperbola x24 − y232 = 1, so Y02 = 8x02-32, | AF | = (x0 − 6) 2 + 8x02 − 32 = 2x0 | 2x0 = 3 (x0 − A2C) {x0 = 2

Please prove that point P (x0, Y0) is in hyperbola ←→ x0 ^ 2 / A ^ 2 - Y0 ^ 2 / b ^ 2 > 1 (including focus)
When the point P is in the hyperbola, the value substituted by the point is greater than 1. This is the opposite of the case where there is a point in the ellipse. Why?

Draw an image, take any point a (x, y) on the hyperbola, make a vertical line to the X axis, let P (x0, Y0) be on the vertical line, then x = x0, from known, x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, that is to say, changing x into x0 is still true, but from known, the absolute value of Y0 must be less than the absolute value of Y, then Y0 ^ 2 < y ^ 2, substituting into hyperbola, we can get x0 ^ 2 / A ^ 2-y0 ^ 2 / b ^ 2 > 1

P (x0, Y0) is a point on the right branch of hyperbola x ^ 2 / a2-y ^ 2 / B2 = 1, then what is the distance from P to the right focus f

If we use the focal radius to solve this problem, we can see the result at a glance, but you must not have learned it. Therefore, we use the first definition of conic to deduce the fixed value of the distance difference between known P and point (- C, 0) and (C, 0), 2A root [(x + C) &# 178; + Y & # 178;] - root [(x-C) &# 178; + Y & # 178;] = 2A shift term root [(x + C) &# 178; + Y & # 178;] = 2A shift term root [(x + C) &# # 178; + Y & # 1

Find the linear equation of point P (x0, Y0) parallel to the line ax + by + C = 0

a(x-x0)+b(y-y0)=0