What about straight line parameters? Thank you very much

You don't see if this method works:

OA crosses the origin, let A(X1, Y1), the slope of natural OA is K1=Y1 X1, known OA 丄L, so the slope of L K=-1 K1=-X1 Y1, and the slope of L is known K=Sina Cosa, so the coordinates of A X1, Y1 are related, -X1 Y1=Sina Cosa, that is, X1=-Y1Sina Cosa, you know that A is on L, so there must be a coordinate of A to meet L, substitute X1 into L, and find Y1=sinacosa (cosa 2-sina 2), and from x1=-y1sina cosa, we get x1=sina 2 (cosa 2-sina 2) and that's it.

1) Simplify the equation of l to obtain: y=(sina cosa)*x-1), let the coordinates of point a be (x,kx), oa is perpendicular to l, so k=-cosa sina, i.e. a(x,-cosa sina x), substitute the l equation, x=(sina), so y=kx=-cosa sina x =-cosa sina * sina) =-sina, cosa

So a((sina), sina, cosa).

1. Parametric equations.

In general, in a planar Cartesian coordinate system, if the coordinates x and y of any point on the curve are a function of a variable t: x=f(t) and y=g(t).

And for each allowable value of t, the points (x, y) determined by the system of equations are on this curve, then this equation is called the parametric equation of the curve, and the variable t that is related to the variables x and y is called the parametric variable, referred to as the parameter. Comparatively speaking, the equation that Hu Zhizhi directly gives the relationship between point coordinates is called an ordinary equation.

2. Parametric equations for straight lines.

1. Solution: From the meaning of the question, we can know that the slope of the straight line l is 1 2, let the linear equation be y=k(x+a), and substitute l, 0=1 2*(-2+a) through the point p(-2,0), get a=2, and deduce l as.

y=1/2(x+2)

Simultaneous parabola c: 5y 2 + x - 2y + 1 = 0

The intersection point is x1=2*(,y1=

x2=-2*(,y2=

Then the length of the chord is: [(x1-x2) 2+(y1-y2) 2] 2. Proof: The coordinates of the midpoint of the string are: x=(x1+x2) 2=-2y=(y1+y2) 2=0

This is the p-point coordinates.

That is, it proves that the p-point is the midpoint of the string.

The parametric equation for a straight line can be rewritten.

x-x')/cosa=(y-y'The sina key is the denominator cosa, the two numbers of sina, and what matters is their ratio (i.e. slope k=sina cosa), not themselves! For example, 2 3 = 4 6 = ......

So it's not surprising that the denominator is greater than 1.

x=1+2t,y=2-3t

It can be rewritten as (x-1) 2=(y-2) (-3), and the denominator is 2 and -3, which means that the slope of the line is -3 2

Conversely, there is a parametric equation x=x for the spear of the Russian evil domain'+at,y=y'+bt, which means a straight line.

1. Set the inclination angle to a

tana=sina/cosa=1/2

cosa=2sina

Substitute sin a+cos a=1

sin²a=1/5

sina=√5/5

cosa=2√5/5

So the straight line is.

x=-2+2t√5/5

y=t√5/5

Substitute ct -2+2t 5 5-2t 5 5+1=0t -1=0

t= 1, so chord length = |t1-t2|=2

2、t²-1=0

So t1+t2=0

If the string is AB, the directed line segments PA and PB are equal in size and opposite in direction.

So p is the ab midpoint.

Let the slope k

k=tan(arctan1/2)=1/2

Linear equation: y=1 2(x+2)=x 2+1

x=2y-2

1) (1) generation 5y 2 + x-2y + 1 = 0 to get 5y 2-1 = 0

y1=√5/5

y2=-√5/5

x1=2√5/5-2,x2=-2√5/5-2

Chord length: l = (x1-x2) +y1-y2) =(4 5 5) +2 5 5).

4. The length of the chord is 22) midpoint abscissa x0 = (x1 + x2) 2 = -2 midpoint abscissa x0 = (y1 + y2) 2 = 0

p(-2,0) is the midpoint of the string.

The inclination angle of 1 is arctan1 2

Explanation: k=tan(arctan1 2)=1 2, so the linear equation x-2y+2=0

The parabola of the generation gets 5y 2 + 2y - 2-2y + 1 = 5y 2-1 = 0y1 = root number 5 5

y2 = - root number 5 5

Chord length = (y1-y2) * root number (1 + 1 k 2) = 22 The midpoint y coordinate is (y1+y2) 2=0

The x-2y+2=0 of the generation is -2, and the coordinate of x is -2

i.e. p(-2,0) is the required midpoint.

1) Let the slope of the oak segment of OA be K, and try K to represent the coordinates of the point AB. Answer A(2P K 2,2P K).b(2pk 2,-2pk) in detail: a

On the parabola y 2 = 2 px, a(x, root number 2pk) straight oa: y=k*x

On the straight beam Bi OA again, the root number is 2PK=K*X

This yields x=2p k2 (2p divided by k squared).

then y=2p k

a(2p/k^2,2p/k).

Two strings perpendicular to each other oa, ob

Straight line ob:y=(-1 k)*x

The same understanding is b(2pk 2,-2pk).

2) Find the trajectory equation for the midpoint M of the chord AB.

b. Absolute value of mo = absolute value of 1 2ab.

Using these two points, it can be found.

Except for one point, m is not in x

axis, otherwise k

does not exist.

: The equation for the straight line l is y=4 3 *(x 2).

Simultaneous equation y 4 3 *(x 2).

y2 2x gives 8x2-41x+32=0

Let a(x1,y1)b(x2,y2) m(x0,y0), then x1+x2 41 8 ,x1x2 4,y1+y2 4 3 (x1+x2 4)=3 2

1）x0＝x1+x2 2 =41 /16 ，y0＝y1+y2 2 =3/ 4

The distance between p,m is pm = (2 41 16 )2 + (0 3 4 )2 15 16

2) The coordinates of the m point (41 16 ,3 4) can be obtained from (1).

3）ab^2＝ (x1−x2)^2+(y1−y2)^2 = (1+16/ 9 )[x1+x2)^2−4x1x2]

So ab = 5 root number 73 8

If there is anything you don't understand, you can ask, I hope it will help you! We hope you will adopt it!

According to the meaning of the problem, the equation of the straight line l can be y=4 3(x-2), and it can be substituted into the parabolic equation y2=2x.

8x∧2-41x+32=0

Let a(x1,y1)b(x2,y2) m(x0,y0) so that x1+x2 41 8,x1x2 4,y1+y2 4 3 (x1+x2 4)=3 2

According to the midpoint formula.

x0=（x1+x2）/2 =41/16，y0=（y1+y2）/2 =3/4

pm｜= √(2−41/16 )∧2+(0−3/4 )∧2 ＝15/16

From , it can be seen that the coordinates of point m are (41 16,3 4) ab = (x1 x2) 2+(y1 y2) 2= (1+16 9)[(x1+x2) 2 4x1x2]= 25 9 (41 2 64 16 )

Your question is not accurate, the first problem has nothing to do with curves, and the second you say the distance has to do with the form of the parametric equation of the straight line. The exact problem is described as follows: (1) The parametric equation of the straight line is known:

x=x0+at where a 2+b 2=1, t is a real number, and y=y0+bt (2) has two points m1(t1) and m2(t2) on a straight line, then |m1m2| =t1-t2|Proof: |m1m2|^2=(x1-x2)^2+(y1-y2)^2 =(at1-at2)^2+(bt1-bt2)^2 =(a^2+b^2)(t1-t2)^2 =(t1-t2)^2 |m1m2|=|t1-t2|Note how to understand: 1. The above known straight lines refer to the equation of Sun Ye's number:

x0,y0) is the fixed point on the line, (a,b) is the unit direction vector of the line, t is the distance from the moving point on the line to the fixed point (x0,y0). 2. If (a,b) is the direction vector of the straight line, but not the unit direction vector, can you classify the parametric equation as the above form?

Let's talk about it:

t=(x-1)/2 、t=(2-y)/3=> (x-1)/2=(2-y)/3

3(x-1)=2(2-y)

3x+2y-7=0

In general, as long as it is a [definite] "straight line equation", then x, y, and t should be (in most cases) x, y, and t all once (e.g. in this problem, but there should still be some [non-)one-time relations that are also [straight] equations).

If the linear (parameter) equations x=f(t) and y=g(t) are to be reduced to regular equations.

Usually the parametric equation can be [solved] t=f (x)=g (y), and then the linear equation u(x,y)=0 can be deduced from the (one-time equation) f (x)=g (y).

x=2+(2 root number 20).

Root number 20) t

y=1+(4 roots, number, 20).

Root number 20) t

then (root number 20) t

Represents the distance between two points on a straight line.

Note that the sum of squares of (2 root number 20) and (4 root number 20) is 1, which is equivalent to cos and sin