Sophomore math parabolic problem, solve 10

1. Proof : Let x=my+n, then m=1 k, n=-b k substitute y 2=36x.

y2-36my-36n=0

y1+y2=36m

y1y2=-36n

y1-y2|^2=36^2*m2+144n=a21296/k2-144b/k=a2

To simplify: a 2*k 2=144(9-kb) (inconsistent with the answer, the question is wrong, it is really harmful).

2. Proof: (Y1+Y2) 2=18M

324m2=36x

x=9m2d(9m2,18m), that is, d(9 k2,18 k) point d to the straight line y=kx+b distance d=|b-9/k|/√(1+k2)s=1/2a*|b-9/k|/√(1+k2)s2 =a2/4|9-kb|2/k2(1+k2)s2 =a4/576

The point shortest to the focal point is the intersection of the parabola and the x-axis.

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There are two types of parabolic equations, one is a quadratic equation with an expression such as y=ax 2+bx+c, where a≠0, and the other is a parabolic shape such as y 2=2px

To solve parabola-related exercises, you need to use the relevant knowledge of parabola.

Such as the symmetry axis, monotonicity, opening and other properties of the parabola, as well as the vertex, alignment, focus and other knowledge content of the parabola.

I've been writing for a long time, and the formula is very difficult to write, so choose me

This is an example question in the book, and it is very simple. Look at the book.

Let y = kx and focus f(k 4,0) straight line: y=k scramble 4-x 16x -24kx+k =0 Let the intersection points be a(x1,y1), b(x2,y2) x1x2=k 16,x1+x2=24k 16=3k 2 ab= (1+k )*x1+x2) reaspertoire-4x1x2]=8 Bring in the above solution, that is, the mu can be absolutely can.

Parabola y = 2x

Vertex (0,0), Focus (1 2,0).

Alignment equation: x=-1 2

Translate the parabola y = 2x to the left by 1 2 units to get the parabola: y = 2[x + (1 2)].

This parabola:

vertices (-1 2, 0).

Focus (0,0).

Alignment equation: x=-1

First of all, you can try to draw a picture, the general image should always be clear.

The vertex is (-1, 2,0).

As for this function, it's not standard, because there's a 1 2, but you can go back to the definition.

What is the focus of the parabola and what is the alignment of the parabola.

If a curve is equal to the distance from any point to a certain point and a certain straight line, then the curve is a parabola, this fixed point is called the focus of the parabola, and the fixed line is called the alignment of the parabola.

Conversely, the distance from any point on the parabola to the focal point and to the alignment is equal

You can set the focus to (m,0) and the alignment equation to x = -m-1

Let the points on the parabola (a, b) be given an equation based on the fact that the distance from the point on the parabola to the focal point is equal to the distance from the point to the alignment. The value of m should have nothing to do with the value of the point on the parabola, so the terms with a and b are omitted. Solve again.

But this is just a personal train of thought, and I'm not sure if it's the right thing to do, because it's the same sophomore in high school. I haven't solved this equation either. It's just that if you think it makes sense, then take it. I affirmed, not necessarily right.

y²=2(x+1/2).The image can be seen as translating the abscissa of all points on the image with y = 2x to the left by 1 2 units.

The vertices of y = 2x are (0,0) focal point (1 2,0), and the alignment x = -1 4, so the corresponding translation must be 1 2 units to the left.

Therefore, the vertices of y = 2 (x + 1 2) are (-1 2, 0) focal point (0, 0), and the alignment x = -1

Let the straight line where cd is located be y=x+b, and the equation of the parabola is connected to get x-square (2b-1)x b-square 0Let the two roots be x1 and x2, and x1·x2 and x1+x2 are obtained from Vedica's theoremThe cube of the Cd equation (y1-y2) is equal to the (x1-x2) square, and the cd distance is (x2-x1) (absolute) of 2 times the root.

x2-x1=root[(x1 x2)square-4x1·x2].The distance between Cd and Ab is 2 times the root (B-4) (absolute), which is equal to Cd. Calculate b and substitute it.

(1) Y = plus or minus root number 3 (x-1).

2) k does not exist.

Application Knowledge Points:

1) The length of the chord over the focus is:

x1+x2+p

2) Vector inner product = 0, then OA is perpendicular to OB

x1x2+y1y2=0

3) Vedic theorem.

You can be specific about the problems of the parabola, and you can solve them according to the outline of the questions.