Math problems about parabolas, math parabolic problems

1. Let y=ax 2+bx+c

c=-216a-4b+c=0

a+b+c=0

a= b= c=-2

y=2, the two triangles are of the same height.

So the ratio of their bottom edge is bf cf=1 2

bf/bc=1/3=be/ab

It is not difficult to see the third equinox of point AB at point E.

The coordinates of point e are (-2 3 0).

3. The last question is to make full use of the condition that PQ is parallel to the y-axis.

That is, their abscissa is equal.

Let's set the abscissa to a

First, find the equation of the AC line and find the ordinate of the q point.

Then use the parabolic equation to find the ordinate of point p.

Their distance is the difference in their ordinates.

If you want to subtract it, you can find the best value.

The equation for the ac line is set to y=kx+b

It's easy to see y=

Then the Q-point coordinates are (a

The coordinates of point p are (a

a is between [-4 0].

Pq length = Q point ordinate - P point ordinate =

0a^2-4a=0

a+2)^2+4=0

When a=-2, pq is the longest = 4

Finding the coordinates is simple.

Construct a function about the coordinates of p and find the maximum.

Obviously, the coordinates of point f are (1,0).

Let a(x1,y1) b(x2,y2) c(x3,y3) (all numbers are subscripts).

fa（x1-1,y1） fb（x2-1,y2) fc(x3-1,y3)

The vector fa plus the vector fb plus the vector vector fc is equal to 0(x1-1, y1) + (x2-1, y2) + (x3-1, y3) = 0, so x1-1+x2-1+x3-1=0x1+x2+x3=3

y1+y2+y3=0

fa+fb+fc=x1+1+x2+1+x3+1 (the distance from the point to the intersection is equal to the distance from the point to the alignment).

x1+x2+x3+3

Have you ever learned to seek guidance? y=(3/8)x2

When the slope of the straight line is the same as the slope of a point on the parabola, the distance is the shortest, and the slope of the straight line is -3 4, and the parabolic equation is y=ax2, and the derivative of the parabola is y=2ax=-3 4, so a=-3 8

It is very convenient to use polar coordinates for this problem: let the distance from f to the alignment be p, the distance from m to the alignment is d, and let f be the pole and the polar axis is to the right |fm|=r

Then d=p+rcos60° due to r d=e=1 r=p+r 2

So r=2p i.e. |fm|=2p

Solution: (1) Let a(x1,y1), b(x2,y2), x1 x2 (point a is on the right side of point b).

Substituting y=kx+2 into y=2x, it is sorted out.

2x²-kx-2=0

x1+x2=k/2，x1x2=-1.

m is the midpoint of the line segment ab, the abscissa of m is (x1+x2) 2=k 4, and mn x-axis.

The abscissa of n is k 4

Derivative of the function y=2x yields y'=4x

So, the tangent slope k of the parabola at n points'=4×k/4=k

Therefore, the tangent slope of the parabola c at point n is equal to the slope of ab.

i.e. the tangent of the parabola c at the point n is parallel to ab

2) Suppose there is such a k

Let n(x0,y0), from the first question x0=k 4, y0=2x0 =k 8

Vector na·vector nb=(x1-x0,y1-y0)· (x2-x0，y2-y0)=0

x1-x0)(x2-x0)+(y1-y0)(y2-y0)=0...

y1y2=2x1 ·2x2 =4(x1x2) =4,y1+y2=2x1 +2x2 =2(x1+x2) -4x1x2=(k 2)+4

Therefore, the formula can be organized as:

k^4+12k²-64=0

The solution yields k = 4 or k = -16 (round).

Therefore, there is k = 2 to satisfy the topic.

I've written down your serial number, and I'll make it for you when I adopt it.

1m has the highest point, B is 5m high, and the distance from the straight line where OA is located is 4m, the radius of the circle is 9m, and the radius of B is 9m, and the vertical line of B to the ground is intersected with D, then the OD length is 4m, 9-4=5m

It is also 5m high, and the two sides are equal and are equilateral right triangles.

The solvable OA is 1M

x=-2a/b=4

y=c-b²/4a=5

Bring (9,0) into y=ax bx c

Three equations are solved.

1.Find the point a(1.)-3), point b (0.-1), c().

Let y=ax 2+bx+c

then a+b+c=-3

c=-14a-2b+c=9

Solution: a=1, b=-3, c=-1

So rock blind y=x 2-3x-1

2.The vertex coordinates of an image with a known quadratic function are (6-12), and pass through the point (, find its relation.

Let y=a(x-6) 2-12

Substituting the point (8,0) yields: coarse scum empty a=3

So y=3(x-6) 2-12=3x 2-36x+963It is known that the quadratic function y=2x2+5x+5It is converted into the form of y=a(x-h)2+k by the matching method, and its vertex coordinates and symmetry axes are written.

y=2x^2+5x+5

2 (x 2+5 beam head 2x) + 5

2(x+5/4)^2+15/8

So the vertices are (5 4, 15 8) and the axis of symmetry is the straight line x=-5 4

1. Parametric equations. If the slope of OA is denoted by k, then the coordinates of a and b, the slope of ab, and the slope of om can all be represented by k, and the parametric equation of m is obtained. 2. Meet the requirements of the question of the straight line ab, constant (2p, 0), this question is the point (2, 0), consider it, you will have other solutions.

Answer: (x 1) 2 y 2=1(x≠0).

Let the point a be (y0 2 2,y0), the slope of the straight line oa can be obtained, and the slope of ob can be obtained by oa perpendicular to ob, and the equation of the straight line ob can be obtained, and the coordinates of point b can be obtained by combining the parabolic equation. The slope and equation of the straight line AB equation and OM are obtained from the coordinates of points A and B, and the results of solving point M by combining the OM and AB equations can be obtained.