Question about the tangent equation of a point on a parabola!

to the parabolic equation.

About x derivative yy'=p, (uses an implicit function.)

derivative), i.e. y'=p/y

Tangent equations. y-y0=y'(x-x0) i.e. y-yo=p y*(x-x0) simplification gives y0y=p(x+x0).

Point of tangency. Chord equation: The slope of the derivative of the tangent point = the slope of the line connecting the two points.

y'=(y-yo)/(x-x0)

Bring in y'=y p, simplifying y0y=p(x+x0).

For a given point p and a given parabola c, if a string ab on c crosses point p and is bisected by point p, then the string ab is said to be the midpoint chord on parabola c that passes point p.

p is the midpoint of ab.

Proof: It is only necessary to prove that the slope of the midpoint chord is also p y, and the rest of the process is the same as above.

Let the straight line where the chord ab is located x-x0=m(y-y0) where m is the reciprocal of the slope, and let m be to avoid discussing the situation where the slope does not exist. Substituting the parabolic equation yields y 2-2pmy+2pmy0-2px0=0

Midpoint So y1+y2= 2pm=2y0

i.e. m=y0 p 1 m=p y0 proves that the slope of the midpoint chord is also p y

Specific questions below.

Question 3: Of course, it can be written like this, in which case the slope of the derivative is y'=x/p

Problem 4: It does not contradict the push wheel 1, the equation is not the same as y 2=2px This is x 2=2py

Question 5: You can write it down as a conclusion, but remember to distinguish between the types of equations.

The question is so long

The first diagram should introduce the three definitions, their definition equations are in the same form, but (x0, y0) are not the same point, and the proof should be calculated slowly according to the derivative. The key string is also a definition outside of the knowledge point, and you can draw it according to the definition with patience. The second picture, I don't see what you mean, what is inference 1?

For parabola.

y 2 2px, the tangent equation of points a(x1,y1) and b(x2,y2) on the parabola.

They are: y1y p(x x1), y2y p(x x2).

Point m(x0,y0) on y1y p(x x1), y1y0 p(x0 x1).·

Point m(x0,y0) on y2y p(x x2), y2y0 p(x0 x2).·

From , we can see that the points a(x1,y1) and b(x2,y2) are on the straight line y0y p(x x0), and the equation for ab is: y0y p(x x0).

Over the parabola y 2 2px a little outside m (feast on this silver x0, y0) as its two tangents, tangents.

The equation for the string is y0y p(x x0).

A: For example, a parabola.

y=x^2+3

One point outside the parabola is (1,1).

Set the tangent point on the parabola.

is (a, a 2+3).

Derivative of the parabola: y'Cong dou finger (x) = 2x

So: cut the congratulatory line.

Slope k=y'(a)=2a =(a 2+3-1) (a-1) so: infiltrate a 2 + 2 = 2a 2-2a

So: a 2-2a = 2

So: (a-1) 2=3

Solution: a=1+3 or a=1-3

So: k=2a=2+2 3 or 2-2 3

So: the tangent is y-1=k(x-1), and it can be substituted.

Parabolic slow sensitization line y x 2 At any point, the tangent of the book is disturbed by the macro slope k=2x, x=1 is substituted to get k=2, so the tangent equation at the point (1,1) is y-1=2(x-1), which is simplified to obtain y=2x-1

<>1. If the equation for the parabola is.

Tap 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: <>

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 <>

Tap 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 <>

Tap 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.

The tangent equation for finding the function can be found like this:

1. Find the derivative of the function first, then there is.

y'=(2-x²)'2x

2. Find the slope at the passing point m(1,1), then there is.

k=y'(1,1)=-2×1=-2

3. Find the tangent equation of the group shouting at the point m(1,1), then the oblique equation of the root collapse field stronghold can be obtained.

y-1=-2(x-1)

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

Teach you an easy and fast way:

1.Find the distance from this point to the focal point (you can use the distance between two points formula, or you can use the distance to the alignment indirectly, in short, the computation of the first step can be negligible).

2.Find a point on the axis of symmetry of the parabola so that the distance from this point to the focal point is equal to the distance obtained in step 1 (there are two such points, take the point outside the parabola).

3.Find the straight line of the known point and the point you obtained in the second step, this straight line is the tangent you are looking for, the principle of this method actually uses the optical properties of the parabola, that is: through any point on the parabola a, the perpendicular line of the alignment, the perpendicular foot is b, connecting a and the focal point f, then the tangent of a is the bisector of the angle baf.

Use the derivative because the definition of the derivative is the slope of a point.

Set a certain point, first find the derivative of the parabola, and bring the abscissa of the point into the tangent slope, and use the point oblique formula.

High school mathematics, analytic geometry, finding tangent equations for parabolas, and thinking from a different perspective.

If you have learned to derive, it is simple.

For example, y=ax +bx+c,y'=2ax+b

The tangent of the crossing point (p,q) is y=(2ap+b)(x-p)+qIf you have not learned to find the derivative, then let the tangent of the crossing point (p,q) be y=k(x-p)+q and substitute it into the parabolic equation to obtain the one-dimensional quadratic equation about x, so that the discriminant formula =0, and kThat is, the tangent is obtained.