Problem description:

1. The square of the square of hyperbola A minus the square of b equals 1, (A >；; 0, b>0) and perpendicular to the hyperbola and intersecting at two points m and n, and the circles with diameters m and n just pass through the right vertex of the hyperbola, then the eccentricity of the hyperbola is _ _ _ _ _.

2. Let the square of hyperbola A minus the square of b be equal to 1, the right focus of (a>0, b>0) is F, and the right directrix 1 intersects with two asymptotes at points P and Q.. If the triangle PQF is a right triangle, the eccentricity of hyperbola is _ _ _ _ _.

3. It is known that the square of an ellipse of 25 x plus the square of 16y is equal to the square of 1 hyperbolic m minus the square of n is equal to 1, (m >；; 0, n>0) has the same focus F 1, F2. Let one focus of two curves be q and the angle qf1F2 = 90, then the eccentricity of hyperbola is _ _ _ _ _.

4. the point p is a hyperbola c 1: the square of x minus the square of b equals to1,(A >；; 0, b>0) and the square of circle C2: the square of x plus y is equal to an intersection point of the square of a plus the square of b, and the angle pF 1 F2 twice = the angle PF2F 1, where f1and F2 are the two focuses of hyperbola C 1, then hyperbola c/kloc.

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Analysis:

Solution:

1.

X=-c is substituted into the elliptic equation, y = b 2/a, and this y value is also the radius of the circle.

Because the circle passes through the right vertex, another expression of radius =a+c is obtained, and the two expressions are simultaneous.

B 2/A = A+C, and because B 2 = C 2-A 2.

Sort out the above formula and get 2 * a 2+a * c-c 2 = 0.

Divide both sides by a 2 at the same time and you can get E 2-E-2 = 0.

The solution is e=2.

2.

Substitute asymptote y=(b/a)*x and right directrix x = a 2/c into asymptote equation, and the ordinate of intersection point p and q is AB/c respectively.

The intersection of PQ and X axis is set to m, and in the right triangle PFQ, PF=QF, so it is an isosceles right triangle, so FM=PM.

C-a 2/c = ab/c = b 2/c, so there is a=b, so e= root number 2.

3.

Elliptic focus (3,0)

Because the angle QF 1F2=90 degrees, QF 1⊥F 1F2.

So q is the intersection of a straight line perpendicular to the X axis and passing through the focus of the ellipse and the ellipse. When x=3, QF 1 is the ordinate = 16/5.

|F2Q|^2+|F 1Q|^2+(2c)^2

|F2Q|=34/5

It is defined by hyperbola | f2q |-| f1q | = 2 * m =18/5, so m=9/5.

e=c/m=5/3

4.

Draw hyperbola, rectangle and asymptote with long axis and short axis as sides respectively, and mark the focus.

The radius square of circle C2 is = a 2+b 2 = c 2, so the circle is centered on the focal line F 1F2, and the triangle PF 1F2 is a right triangle.

According to the meaning of the question, we can get that the angle PF 1F2 is 30 degrees, PF2=( 1/2)*(2*c)=c, PF 1= root number three *c.

Defined by hyperbola | pf1| | pf2 | = 2 * a = (root number 3-1)*c

E=c/a= root number three+1.

Note: Q in the third question is the non-focus of intersection.