Given that the ratio of the distance between the moving point P and the fixed point F (2,0) and its distance to the fixed line L: x = 8 is 1:2, the trajectory equation of P is obtained

Given that the ratio of the distance between the moving point P and the fixed point F (2,0) and its distance to the fixed line L: x = 8 is 1:2, the trajectory equation of P is obtained

Let P point coordinate be (x, y), then the distance from P to f is sqrt [(x - 2) ^ 2 - y ^ 2]
The distance to the straight line is | X-8|
From the meaning of the title, we can see that sqrt [(x - 2) ^ 2 - y ^ 2] = 2|x-8|
That is, (X-2) ^ 2-y ^ 2 = 4 (X-8) ^ 2
We can get the trajectory equation

If the line L passes through the point (0,2), and passes through the intersection of two lines 2x minus 3Y minus 3 equal to 0 and x plus y plus 2 equal to 0, we can find the solution of the equation of the line L
If the line L passes through the point (0,2), and passes through the intersection of two lines 2x minus 3Y minus 3 equal to 0 and x plus y plus 2 equal to 0, we can find the solution of the equation of the line L

Solve the equations of 2x-3y-3 = 0 and X + y + 2 = 0, find out the focus (- 3 / 5, - 7 / 5), and then use the known points (0,2), use the formula to find out the linear equation, [2 - (- 7 / 5)] / [0 - (- 3 / 5)] = 17 / 3 is the linear slope
y-2=17x/3

If three straight lines 2x + 3Y + 8 are equal to 0, x minus y minus 1 is equal to zero, and x plus K, y is equal to zero, the value of K is obtained at a point of intersection

K=-1/2
In the upper right corner of my answer, click [adopt answer]

The perimeter of a right triangle is 2. What is its maximum area?
What is the maximum area of a right triangle if its perimeter is 2?

Let the right angle sides be x and Y respectively, then the hypotenuse is under the root sign (x ^ 2 + y ^ 2)
Area s = XY / 2
From the mean inequality
2 = x + y + radical (x ^ 2 + y ^ 2) > = 2 * radical (XY) + radical (2XY)
So the radical (XY) = 2 / (2 + radical 2) = 2-radical 2
S = 3-2 * radical 2

If the perimeter of a right triangle is (2) + 1, what is its maximum area?

Let: one right angle side be x, the other right angle side be y, hypotenuse = root sign (x ^ 2 + y ^ 2) x + y + root sign (x ^ 2 + y ^ 2) = root sign 2 + 1. This equation is too boring. We still think that a right triangle with the largest area must be isosceles

2. If the triangle ABC is a right triangle, the two right sides are 5,12, and P is a point in the triangle ABC,
The distance from it to the three sides of the right triangle is equal, and the distance is ()

Let this point be o, then o is the center of the triangle ABC, that is, the center of the inscribed circle. The right angle sides of the triangle ABC AB = 12, BC = 5, then AC = 13 is the line from O to the vertical sides, OC (on AB side), OD (on BC side), OE (on AC side) let AC = x, so CB = BD = 12-x, DC = EC = 5 - (12-x) = X-7, then AE = ac-ec = 13 - (X-7) = 20

If the perimeter of a right triangle is 2, what is its maximum area?

Answer: 3-2 √ 2
Let the hypotenuse be c,
a=csinα,
b=ccosα,
a＋b＋c=2,
c(1＋sinα＋cosα)=2,
c[1＋√2 sin(α＋∏/4 )]=2,
c≤2/（1+√2） =2(√2 －1),
S△= c^2sin2α≤1/4c^2=3－2√2

The sum of two right angles of an isosceles right triangle is 8.8 decimeters. How many square decimeters is its area?

It should be 9.68

An isosceles right triangle two right angles side suitable 16 decimeters, its area is how many square decimeters?

16 △ 2 = 8 decimeter right angle side
8 × 8 △ 2 = 32 square decimeter area

The right side of an isosceles right triangle is 2.5dm. How many square decimeters is its area?

Because it is an isosceles right triangle, the two equal angles are 45 degrees, and the height of the triangle is 1.25 decimeters,
Finally, 1.25 × 2.5 △ 2 = 1.5625 square decimeter
This is the calculation of the hypotenuse, and the right angle side is their correct calculation