A high school math analytic geometry problem to help me solve a high school math analytic geometry p

Solution: Establish a plane Cartesian coordinate system through point d, let the radius of the circle be r, and the coordinates of point c be (0,y), according to the similar triangle theorem, there is.

y (2- 2) = [(y-r) -r ] r, squared on both sides, get.

y²/(6-4√2)=[(y-r)²-r²]/r²，r²y²=(6-4√2)(y²-2ry)，∵y>0，r²y=(6-4√2)(y-2r)，y=(12-8√2)r/(6-4√2-r²)

It can be seen that it is a straight line parallel to the x-axis.

x2-2x+2y2-1=0 The main thing is to find a relationship, ABC forms a triangle, uses the same point to draw two tangent lines with equal lengths, knows that AC length minus BC length is equal to 2 root number 2, and then set c to (x, y), and the column equation is simplified.

For the sake of clarity, vector symbols are omitted.

Solution: (1) is given by a(0,1), b(0,-1), c(1,0), and p(x,y).

then the vector op(x,y);

ap（x-1，y）；

bp（x+1，y）；

pc（1-x，-y）；

by ap*bp=k|pc|2. Substitution:

x+1)(x-1)+y^2=k[(1-x)^2+(-y)^2];

Solution: k=1;x=1；

Substituting yields p(1,y).

That is, the trajectory of the point p is a straight line x=1;

2) by (1) in ap(x-1,y); bp（x+1，y）；

Substitution|2ap+bp|Gotta :

2ap+bp|=|(3x-1,3y)|=[（3x-1）^2+9y^2]^(1/2)

When k=2, ap*bp=k|in (1) is substitutedpc|2. Got:

x-2）^2+y^2=1;

Substituting it into the above equation yields:

2ap+bp|=(30x-26)^(1/2)

Since (x-2) 2+y 2=1; The geometric meaning is that if the center of the circle is (2,0) and the radius r=1, then the range of its x is [1,3];

Substitute x=1 and x=3 into |, respectively2ap+bp|=(30x-26) (1 2), get:

max=8；

min=2；Solution.

Let the slopes of these two lines, one of them be k, and since they are perpendicular to each other, the other will be -1 k.

Let the coordinates of point p be (x. ，y。) then the straight line l1 is y-y. =k(x-x。)

The straight line l2 is y-y. =-1/k(x-x。Lianli L1 L2 can use Y with Y.

Then, the expressed y value is directly substituted into the elliptic equation, and the result obtained is the requirement, and then the y in the resulting equation is used. Replace with y, x. Switching to x is the answer.

Very simple topic:

Let the coordinates of point p be (x0,y0).

The tangent between the p point and the ellipse is (y-y0)=k(x-x0) (k and 1 k exist), then the other tangent is (y-y0)=-1 k(x-x0) respectively substituted into the elliptic equation, take δ=0, eliminate k, and the remaining relationship between y0 and x0 is the equation, and k=0 is substituted, that is, (a, b) whether the four points meet the equation.

Knowing the fixed point p, the point slope of l1 can be y=k1(x-1)+3 2, l2; y=k2(x-1)+3 2 respectively bring the two equations into the ellipse, obtain the system of equations about k1 and k2, combine k1k2=-3 4, solve k1k2, solve ab, simplify (2) solve the length of ab with the two-point formula, and then use the distance formula from the point to the straight line to solve the height of pab, obtain the triangle area function about x, and find x when the function takes the maximum value

I'm not very good at simple methods, stupid should be able to do it, you try, both linear equations know how to synthesize with the ellipse, replace y, and solve it, with k, reuse -3 4, about the same.

1.Set the point m(x1,y1).

x1-16/5| 4

y²+（x-5² 5

Simplification: 9x1 -16y1 = 16*9

i.e. x 16-y 9=1(x 16 5)2It can be seen that A and B are two foci.

Let pa be x and pb be y

RT PAB, y-x=2a=8

Pythagorean theorem: x +y = 100

x²+y²-2xy=64

xy=18, i.e. |pa|*|pb|=18

Solution: The focal point of the parabola is f(a,0).

Let p(x1,y1),q(x2,y2).

Then: (y1) 2=4ax1,y2) 2=4ax2

Subtract, and factor:

y1+y2)(y1-y2)=4a(x1-x2)deformation:(y1-y2) (x1-x2)=4a (y1+y2)note that the slope of pq k=(y1-y2) (x1-x2) is obtained from the above equation: k=4a (y1+y2) (1) and the vector pf=(a-x1,-y1).

fq=(x2-a,y2)

From PF=2FQ, a-x1=2(x2-a)-y1=2y2

That is, x1=3a-2x2 *

y1=-2y2 *

In this way, (1) becomes k=4a (-y2)=-4a y2 (2) and the slope of k=fq =(0-y2) (a-x2) (3) is obtained from (2) and (3).

4a/y2=-y2/(a-x2)

i.e. (y2) 2=4a(a-x2).

i.e. 4a*(x2)=4a(a-x2) (curve equation (y2) 2=4ax2).

i.e. there is (x2) = a 2

Thus: (y2) 2=4a(a 2)=2a 2y2=(root number 2)*a, or the slope of y2=-(root number 2)apq k=2*(root number 2).

or k = -2 * (root number 2).

Let m r, in a planar Cartesian coordinate system, the vector a=(x+ 3,my), the vector b=(x- 3,y), the vector a vector b, and the trajectory of the moving point m(x,y) is the curve eQ: given m=3 4,f(0,-1), and the straight line l:y=kx+1 intersects the curve e at two different points m, n, then is there a maximum value for the area of the inscribed circle of fmn?

If it exists, find the maximum value and the value of the real number k at this time; If not, state the reason.

Analysis: vector a=(x+3,my), vector b=(x-3,y), vector a vector b (m r).

Vector a·vector b = x 2-3 + my 2 = 0

x^2/3+y^2/(3/m)=1

m=3 4x 2 3+y 2 4=1, and the curve e is an ellipse focused on the y-axis.

c = 1f (0, -1) is the lower focus of the curve e.

The straight line y=kx+1 intersects the curve e at two different points m, n

y^2=k^2x^2+2kx+1

Substituting the ellipse yields (4+3k2) x 2+6kx-9=0

From Veda's theorem, x1+x2=-6k (4+3k2), x1x2=-9 (4+3k2).

x1-x2|=√⊿/(4+3k^2)=12√(k^2+1)/(4+3k^2)

s(⊿fmn)=1/2*2*|x1-x2|

Let f(k) = 12 (k 2+1) (4+3k 2).

When k=0, the function f(k) takes the maximum value of 3

Obviously, when the FMN area is the largest, the inscribed circle area is also the largest.

That is, the straight line l is y=1,m(3 2,1),n(-3 2,1),f(0,-1).

fm|=|fn|=5/2，|mn|=3

Let s=1 2(5 2+5 2+3)=4

The radius of its inscribed circle r=s s=3 4

Inscribed circle area = r 2 = 9 16 and k = 0 at this time

e: x^2-3 =3/4y^2 x^2/3 - y^2/4 = 1;It's hyperbola, and the meaning in the question seems that f is the focus, is there a mistake in the data, m = -3 4???Please check it out.

Answer to the second question:

Vector f1m=(x0+2,y0).

Vector f1a + vector f1b + vector f1o = (x1 + x2 + 6, y1 + y2) so x0 = x1 + x2 + 4

y0=y1+y2

Let the straight line of f2 be y=k(x-2) and substitute it into x-squared-y-squared=2

It can be obtained by using Weida's theorem.

x1+x2=(4*k) k-1x1x2=(4k's squared+2) k's square-1, so x0=(8k's squared-4) k's square-1··· Formula.

y0 = square of k (x1x2-2x1-2x2+4) = (square of 2k) 1-k squared··· Two-form.

Simultaneous two formulas can obtain x+2y-4=0

I made it last night.

I don't know if it's right, hehe, it's so hard