Get f (x) = (1/sin2x)+(11-sin2x) =1/sin2x+1/cos2x.
& lt=[( 1/sinx+ 1/cosx)/2]^2
The equal sign holds if and only if sinx=cosx, that is, sinx= 1/ radical number 2, that is, sin 2x = 0.5, that is, original x = 0.5, then the minimum value of the solution is 1.
the second question
(1) circle c: x 2+y 2-6x-8y+2 1 = 0.
That is, (x-3) 2+(y-4) 2 = 4.
L:kx-y-4k+3=0。
That is y=k(x-4)+3.
Therefore, circle C is a circle with center at point (3,4) and radius of 2.
And the straight line c represents all straight lines passing through the fixed points (4, 3) except x=2.
Because (4,3) is in a circle, there must be two intersections.
(2) According to the principle that the longer the chord is, the shorter the chord center distance is (the theorem in the book when learning circle in grade one)
So the straight line passes through point (4,3) and is perpendicular to the straight line passing through points (4,3) and (3,4), so it is obvious that k = 1.
The final linear equation is x-y- 1=0.
Third question
(1) If we set the coordinates of point A (x 1, y 1) and point B (x2, y2), because "the circle with the diameter of AB passes through the origin O", it means:
→ →
Oa ob = 0 (the circumferential angle of the diameter is a right angle)
That is, x 1x2+y 1y2=0.
The linear equation and parabolic equation are simultaneous, ensuring that the tower is replaced >: 0, solving p by Vieta theorem, and finally calculating P = 2.
(2) Set the coordinates of point A (x 1, y 1), point B (x2, y2) and point M (x, y) because.
→ → →
MF=FA+FB
So go ahead.
x- 1 = x 1- 1+x2- 1 x = x 1+x2- 1
y=y 1+y2
According to Vieta's theorem, x = (7k 2+4)/k 2.
Y=4/k, that is, k=4/y is substituted into the above formula.
Find y 2 = 4 (x-7)
Sorry, I can't do the fourth question. . . I've learned it. .
But I guess a =- 1 and b = 0. Either you prove it, but it should be true.
The fifth question
(1) line and hyperbola are simultaneous, so that the coefficient in front of the quadratic term is not 0, and the tower >: 0. It is found that K belongs to (-root number 2,-1) u (- 1, 1)) u (root number 2).
(2) Calculate the base and height of triangle with chord length formula and distance formula from point to straight line respectively, and then substitute them for solution.
According to the chord length formula, [2* root sign (2-k 2) * root sign (1+k 2)]/| 1-k 2 |
The distance from the origin to the straight line L is 1/ root sign (k 2+ 1).
So the area is 1/2*[2* root number (2-k 2) * root number (1+K2)]/|1-K2 | *1/root number (k 2+65438).
Find k 2 = 0 or k 2 = 1.5.
But don't forget that straight lines and hyperbolas intersect at the left and right branches.
Therefore, we found that:
When k 2 = 1.5, x1+x2 > in the quadratic equation of one variable obtained at the same time; 0，x 1 * x2 & gt； 0, because the straight line must pass through the fixed point (0,-1), judging from the image, it shows that the two intersections must be on the same branch, which obviously does not meet the conditions and is discarded.
When k=0, the condition is obviously satisfied.
So in summary, k=0.