The problem of parabola and vector, the derivation of parabolic properties

The focal coordinates are (1 2,0).

The equation for the straight line over the focus is, y=k(x-1 2).

The coordinates of the two points ab can be obtained by concatenation.

y=k(x-1/2)

y^2=2x

The elimination element is converted into two unary quadratic equations.

k^2x^2-k^2x-2x+(1/4)k^2=0y^2-2y/k-1=0

Quadratic equations with solutions can be turned into.

a(x-x1)(x-x2)=0.

So. x1x2=c/a=(1/4)k^2/k^2=1/4y1y2=c/a=-1/1=-1

oa·ob=|oa||ob|cosaob

ab|^2=|oa|^2+|ob|^2-2|oa||ob|cosaob

y1-y2)^2+(x1-x2)^2=(x1^2+y1^2)+(x2^2+y2^2)-2oa·ob

2y1y2-2x1x2=-2oa·ob

So oa·ob=x1x2+y1y2=(1 4)+(1)=-3 4oa·ob=x1x2+y1y2

This is a very important conclusion, and it is important to remember that there are often multiple-choice questions in the college entrance examination.

I remember that there seems to be a person who, because it is a multiple-choice question considering a special situation, has made a straight line with the focus perpendicular to the x-axis, and asked for it.

Looks like there's a right answer?

Wrong topic?

The quasi-old definition of a parabola is a trajectory formed at a point in the plane at the same distance to a fixed point f and a definite straight line (f is not on l). where the point f is the focal point of the parabola and the straight line l is the alignment of the parabola.

Sets the axis of symmetry of the parabola as the coordinate axis and the vertex as the origin. Below we derive the equation based on the nature of the parabola. If the focal coordinates of the parabola are f(p 2,0) and the quasi-line equation is x=-p 2, assuming that the coordinates of any point on the parabola are (x,y), then the following equation holds.

This is the parabolic equation about the x-axis symmetry, with a fixed point at the origin. When p is a positive real number, x is a non-negative value, that is, the opening direction of the parabola is the positive direction of the x-axis; When p is a negative real number, x is a non-positive value, i.e., the opening direction of the parabola is negative on the x-axis.

The most concise way is to do this, because all objects on Earth have a constant acceleration g downward, so for a falling body with mass m there is: a=-g and a=dv dt v=dx dt If the function of motion is known to be y=f(t), then y'=dx dt then y''=dy'dt, so the first equation can be rewritten as: y''=gy''=g

That is to say, when an object falls freely, the distance it falls is proportional to the square of time, and the graph that draws the time distance function is a standard parabola, but you can't see the image parabola in the vertically falling object.

When we throw the object diagonally upwards like a basketball, we will intuitively see that the trajectory of the object presents a parabolic limb with pulsation, according to the above falling law, for the object thrown at an oblique angle, if the velocity when it is thrown is v0.

Simultaneous line and parabolic equation to obtain the coordinates of the m point (3,2 times the root number 3).

It is clear that the triangle nmf is an equilateral triangle with a distance of 2 times the root number 3 from m to nf

Simple questions, practice more and become proficient.

1) Let a(xa,ya), b(xb,xb), c(xc,yc), d(xd,yd), p(x1,y1), q(x2,y2).

From (x-5) 2+y 2=9 and y 2=2px: x 2-2(5-p)x+16=0

So x1=(xa+yb) 2=5-p

y1 = (ya+yb) 2 = root number (2p) 2 * (root number xa + root number xb) = root number (2p) 2 * root number (xa + xb + root number (xaxb)) = root number (2p) 2 * root number (9p-p 2).

From (x-6) 2+y 2=27 and y 2=2px: x 2-2(6-p)x+9=0

So: x2=(xc+xd) 2=6-p

y2 = (yc + yd) 2 = root number (2p) 2 * (root number xc + root number xd) is similar to y1, y2 = root number (9p-p 2).

Then ,|x1-x2|=1

y1-y2|=0

So. pq|=1

2）s△abq=s△apq+s△pbq

pq/2*|ya-yb|

Root number (2p) 2*|Root number xa - root number xb|= root number(p(1-p)).

Because 0, when p=1 2, s abq takes the maximum value of 1 2

The center of the circle is (2,0).

Then find the nearest distance from y 2 = 12 + 4x to the center of the circle.

The minimum value of y 2+((y 2-12) 4-2) 2 is the derivative of the above equation.

2y+solution: y=0, y=32

y=0 does not fit the topic.

When y=32, x=5

The equation for the circle is: (x-2) 2+y 2=9+32

The standard formula for a straight line is: y=ax+b, a is the slope of the straight line, for the straight line of the zero crossing point b=0, the slope is the ordinate of the point on the straight line divided by the abscissa, that is, a=t (t p)=p t, so the straight line on,oh equation is: y=(p t)x

For the length of on,oh, you can use the formula for the length of the line segment between two points d= [(a1-a2) +b1-b2) ] Because o is the origin in the problem, the distance can be expressed as d= (a1 +b1), where a,b are the horizontal and vertical coordinates of the point.

Know the coordinates of h, n, using the formula for the distance between two points.