The derivation process of the parabolic tangent equation

The parabola y = 2px is the conic equation but not a function, the two parts divided by the x axis are functions, and the two corresponding inverse functions together are a function, i.e. y = x (2p), which is also a parabola, and is symmetrical with the parabola y = 2px with respect to the straight line y = x;

Let any point on the parabola y=x (2p) be m(x0,x0 (2p));

From the parabolic image, it can be seen that the tangent of any point on it cannot be parallel to the y-axis, that is, the slope of the tangent at any point on it exists, and the slope of the point past m is k, then the tangent equation is y-(x0 (2p))=k(x-x0);

Simultaneous y=x (2p), subtract y to get: (1 (2p))x -kx+(kx0-(x0 (2p)))=0;

then δ=(-k) -4(1 (2p))(kx0-(x0 (2p)))=0, simplifying k-2(x0 p)k+(x0 p)=0, and obtaining k=x0 p;

For example, to find the tangent equation of y=ax2+bx+c at (x1,y1) point, first derive y=ax2+bx+c on both sides to obtain the tangent slope k=2ax+b, and substitute x1 to obtain it, and then let the tangent equation be y=kx+d, and the equation can be obtained by passing the tangent (x1, y1).

Parabolic tangent equations.

Derivation of the formula: Let the parabola y 2=2px on a point m(x0., y0) is the slope of k, then the tangent equation obtained from the point oblique formula is: y-y0=k(x-x0);

Compare it to a parabolic equation.

Lianli, can get k 2*x 2-2(k 2*x0-ky0+p)x+(y0 2+k 2*x0 2-2k*x0*y0)=0.

Because there is only one slope of the tangent line of the crossing point m, the above equation δ=0, i.e., [-2(k 2*x0-ky0+p)] 2-4k 2*(y0 2+k 2*x0 2-2k*x0*y0)=0;

k=[2y0 (4y0 2-8p*x0) (1 2)] 2*2x0).

Because m(x0.,y0) on the parabola y 2=2px, so y0 2=2px0, substituting the above formula, the brother stove simplifies k=y0 (2x0);

Substituting the point oblique formula, we get y0 2 p*y=y0*(x+x0), that is, y0*y=p(x+x0).

The tangent is sold in the parabola.

As follows: the tangent equation at the point above the parabola y2=2px (x0,y0) is y0y=p(x+x0), and the equation for the slope of the past focus on the parabola y2=2px is k (x-p 2). In-plane refers to the trajectory of the point where the distance from the fixed point is equal to the fixed line.

It's called a parabola. Among them, the fixed point is called the focus of the parabola, and the fixed line is called the quasi-line of the parabola.

The tangent equation for the parabola is as follows: the tangent equation at one point (x0,y0) on the parabola y2=2px is y0y=p(x+x0), and the equation for the slope of the focus over the parabola y2=2px is k=k(x-p 2). In the plane, the distance to the fixed point is equal to the fixed line.

The trajectory of a point is called a parabola. Among them, the fixed point is called the focus of the parabola, and the fixed line is called the quasi-line of the parabola.

The tangent equation for a parabola is:

1. If the equation for the parabola is.

If the point p is on the parabola, then the tangent equation for the parabola passing the point p is:

2. Derivation process:

Let the tangent equation be .

Simultaneous tangent and parabola, simplified to obtain:

Sorted out because the two are tangent, so =0

It can be reinstated:

Parabolic tangent equations.

1. The tangent point Q(x0,y0) is known, and if y 2px, the tangent line y0y p(x0 x); If x 2py, then tangent x0x p (y0 y) and so on.

2. Known tangent points q(x0,y0).

If y 2px, then tangent y0y p(x0 x).

If x 2py, then the tangent x x p(y0 y).

3. The tangent slope k is known

If y 2px, then tangent y kx p (2k).

If x 2py, then the tangent x y k pk 2 (y kx pk 2).

Parabolic geometric properties.

1) Let the tangent of a point p on the parabola intersect with the quasi-line at q, and f is the focus of the parabola, then pf qf. And if P is perpendicular to the alignment and the perpendicular foot is A, then PQ bisects APF.

2) If a point P on the parabola is used as the perpendicular line of the alignment, then the bisector of APF and the parabola are tangent to P. From this property, it can be concluded that the ruler diagram method of the tangent line of the parabola is made by a point p on the parabola.

3) Let the tangent and normal of the parabola point p (p is not the vertex) intersect a and b respectively, then f is the midpoint of ab. This property can be derived from the optical property of the parabola, that is, the rays reflected by the parabola through the focal rays are parallel to the axis of symmetry of the parabola. Various searchlights and automobile lights use this property of parabola (surface) to make the light source at the focal point to emit (quasi-)parallel light.

There is no formula for the tangent equation of a parabola.

The standard parabola is divided into.

y^2=2px

x^2=2py

y^2=-2px

x^2=-2py,p>0

and so on, 3,4 terms are extensions of 1,2 terms.

For the parabolic equation y 2=2px, the tangent of a point m(a,b) on the parabola can be set to y-b=k(x-a).

Simultaneous tangent and parabola.

y=k(x-a)+b

Then [k(x-a)+b] 2-2px=0

K 2x 2-(2K 2A+2P-2Kb)x+K 2A 2+B 2-2kba=0

Obtained by tangentiality. =0, i.e., (2k 2a+2p-2kb) 2-4k 2*(k 2a 2+b 2-2kba)=0

k=p b can be obtained.

Substitution back y-b=k(x-a).

y=p/b*(x-a)+b

In the same way, for the x 2=2py type, the tangent equation y=a p*(x-a)+b can be obtained

--The above is the slope obtained by using the equation synonymous =0.

If you want to learn the derivative, you only need to find the derivative on both sides of the parabolic equation to get the derivative of the change point, that is, the tangent slope, and get the equation.

In addition, the x 2=2py type should note that the slope of the parabolic vertex does not exist, which should be discussed separately.

The tangent equation is related to the parabolic equation and the conditional form of the tangent.

1) The tangent point q(x0,y0) a...If y 2px, then tangent y0y p(x0 x).

b。If x 2py, then the tangent x x p(y0 y).

2) The tangent slope k a is known. If y 2px, then tangent y kx p (2k).

b。If x 2py, then the tangent x y k pk 2 (y kx pk 2).

1) The tangent point q(x0,y0) a...If y 2px, then tangent y0y p(x0 x).

b。If x 2py, then the tangent x x p(y0 y).

2) The tangent slope k a is known. If y 2px, then tangent y kx p (2k).

b。If x 2py, then the tangent x y k pk 2 (y kx pk 2).

The tangent equation for the parabola y=f(x) at the point (x0,y0) is.

y-y0=f(x0,y0)'x-x0

Parabolic tangent equations.

1. The tangent point Q(x0,y0) is known, and if y 2px, the tangent line y0y p(x0 x); If x 2py, then the tangent x0x p (y0 chain empty y) and so on.

2. Known tangent points q(x0,y0).

If y 2px, then tangent y0y p(x0 x).

If x 2py, then the tangent x x p(y0 y).

3. The tangent slope k is known

If y 2px, then tangent y kx p (2k).

If x 2py, then the tangent x y k pk 2 (y kx pk 2).

If the equation for an ellipse is <>

Dots p <>

On an ellipse, then the tangent equation for the ellipse is <>

Proof: The ellipse is <>

The tangent point is <>

then <>

Deriving an ellipse yields <>

That is, the tangent slope is <>

So the tangent equation is <>

Substituting (1) and simplifying the tangent equation is <> if the hyperbolic equation is a disadvantage, it <>

Dots p <>

On hyperbola, then the tangent equation of the hyperbola of the point p is <> the proof of this proposition is similar to that of an ellipse.

There is no formula for the tangent equation of a parabola.

The standard parabola is divided into.

y^2=2px

x^2=2py

y^2=-2px

x^2=-2py,p>0

and so on, 3,4 terms are extensions of 1,2 terms.

For the parabolic equation is y 2=2px, the tangent of one point m(a,b) on the parabola.

The tangent equation can be set to y-b=k(x-a).

Simultaneous tangent and parabola.

y=k(x-a)+b

then <>k(x-a)+b]^2-2px=0

Sorted out. k 2x 2-(2k 2a+2p-2kb)x+k 2a 2+b 2-2kba=0 is obtained by tangent.

i.e. (2k 2a + 2p - 2kb) 2-4k 2*(k 2a 2 + b 2-2kba) = 0

k=p carries the missing b to be obtained.

Substitution back y-b=k(x-a).

y=p/b*(x-a)+b

In the same way, the tangent equation can also be found for the x 2=2py type.

y=a/p*(x-a)+b

--The above is the slope obtained by using the equation synonymous =0.

If you want to learn the derivative, you only need to find the derivative on both sides of the parabolic equation to get the derivative of the change point, that is, the tangent slope, and Xian obtains the equation.

In addition, the x 2=2py type should pay attention to the slope of the parabolic vertice, which should be discussed separately.

The tangent equation for a parabola is:

1. If the equation for the parabola is.

Dots p <>

On a parabola, the tangent equation of the parabola passing through the point p is :

2. Derivation process:

Let the tangent equation be .

Simultaneous tangent and parabola, simplified to obtain:

Sorted out. <>

Because the two are tangent, =0 is indicated

Obtainable. <>

Bring it back to the generation: <>

There are several ways to find a parabola.

1. To find the distance from the point to the focal point, you can use the formula of the late distance between two points, or you can use the distance to the line to obtain it indirectly;

2. Find a point on the axis of symmetry of the parabola so that the distance from this point to the focus is equal to the distance obtained in step 1;

3. Find the straight line of the known point and the point obtained in the second step, which is the tangent line sought;

4. The principle actually uses the optical properties of the parabola, that is, if you pass any point A on the parabola, the perpendicular line of the alignment line, the perpendicular foot is B, and the tangent line of A is the bisector of the angle BAF when connecting A and the focal sail point F.