Find a point P on the ellipse x ^ 2 / 20 + y ^ 2 / 56 = 1, so that the line between point P and two focal points is perpendicular to each other

Find a point P on the ellipse x ^ 2 / 20 + y ^ 2 / 56 = 1, so that the line between point P and two focal points is perpendicular to each other

Easy to find: the two focal points of the ellipse are (0, - 6) and (0,6);
Let the coordinates of point p be (x, y),
Then the equations are obtained
x2/20 + y2/56 = 1 ,
(y+6)/x·(y-6)/x = -1 ；
The solution is as follows
x = ±10/3 ,y = ±(4/3)√14 .
Therefore, there are four possibilities for the coordinates of point P:
（ -10/3 ,-(4/3)√14 ）；
（ -10/3 ,(4/3)√14 ）；
（ 10/3 ,-(4/3)√14 ）；
（ 10/3 ,(4/3)√14 ）.

If the ellipse y225 + X29 = 1 and the hyperbola y215 − x2 = 1 have a common point P, then the area of the triangle formed by P and the hyperbola two focus line is ()
A. 4B. 55C. 5D. 3

According to the definition of ellipse: Pf1 + PF2 = 10, according to the definition of hyperbola: pf1-pf2 = 215, ｜ Pf1 = 5 + 15, PF2 = 5-15, in the triangle pf1f2, F1F2 = 8, according to the cosine theorem: cos ∠ f1pf2 = pf & nbsp; 1 & nbsp; 2 + pf & nbsp; 2 & nbsp; 2 & nbsp; & nbsp; - F & nbsp; 1F & nbsp; & nbsp; 2 & nbsp; 22pf & nbsp; 1pf & nbsp; 2 = 45p and hyperbolic bifocal F1F2 form a triangle with an area of S = 12pf1 · pf2sin ∠ f1pf2 = 12 (5 + 15) (5-15) × 35 = 3, so D is selected

Let the distance from a point (4, t) on the parabola y2 = 2px (P > 0) to the focus be 5.1, and then calculate P and t
2. If the chord length of the straight line y = 2x + B cut by the parabola is 3 root sign 5, find B
3. Find the shortest distance from the moving point m on the parabola to the fixed point a (m, 0)

Let the distance from a point (4, t) on the parabola y & # 178; = 2px (P > 0) to the focus be 5. (1), find P and T; (2), if the chord length of the straight line y = 2x + B cut by the parabola is 3 √ 5, find B; (3), find the shortest distance from the moving point m on the parabola to the fixed point a (m, 0)
(1) Focus (P / 2,0); distance from point (4, t) to focus = √ [(4-P / 2) &# 178; + T & # 178;] = 5, namely:
(4-p/2)²+t²=25.(1)
t²=8p.(2)
(2) Substituting (1) and simplifying, we get P & # 178; + 16p-36 = (P-2) (P + 18) = 0, so we get P = 2, t = 4
(2) By substituting P = 2 into the parabolic equation, we get y & # 178; = 4x. (3)
Then y = 2x + B is substituted into (3) to get: (2x + b) &# 178; = 4x; the expansion is simplified to 4x & # 178; + 4 (B-1) x + B & # 178; = 0
Let the intersection of the straight line and the parabola be a (X &;, Y &;), B (X &;, Y &;), then;
Chord length AB = (√ 5) √ [(X & # 8321; + X & # 8322;) &# 178; - 4x & # 8321; X & # 8322;] = (√ 5) √ [(B-1) &# 178; - B & # 178;] = (√ 5) √ (- 2b + 1) = 3 √ 5
It is reduced to (1-2b) &# 178; = 9, that is, there are 4B & # 178; - 4b-8 = 0, B & # 178; - B-2 = (b-2) (B + 1) = 0, so B & # 8321; = 2 (rounding); B & # 8322; = - 1
(if B = 2, then the line y = 2x + 2 has no intersection with the parabola, because the discriminant Δ of equation 4x & # 178; + 4x + 4 = 0

If the distance a from a point m to the focus on the parabola y2 = 2px (P greater than 0) (a greater than p divided by 2), then the distance from the point m to the collimator is. The abscissa of the point m is. 2 the coordinate of the point on the parabola y2 = 12x whose distance from the focus is equal to 9 is
Please write down the steps to solve the problem,

1. Focus on (P / 2,0)
The guide line is: x = - P / 2
The distance to the guide line is a (parabola eccentricity is 1)
From the above two points
The abscissa is a - (P / 2)
The focal point is (3,0) and the collimator is x = - 3
So the abscissa of such a point is - 3 + 9 = 6
So these points are (6,6 times radical 2) and (6, - 6 times radical 2)