What are the focal coordinates and the Quasilinear equation of the parabola y = 2px?

This is a linear function
Any one has a square and the other doesn't
That's the parabola
If y = 2px
The guide line is x = - P / 2
Focus coordinates (P / 2,0)
If y = 2px
The first deformation is Y / 2p = X
The guide line is y = - 1 / 8p
Focus (0,1 / 8p)

The Quasilinear equation and circle of the known parabola y ^ 2 = 2px (P > 0)

The Quasilinear equation of Y & # 178; = 2px is: x = - P / 2

Let the point (x0, Y0) be the point on the parabola y = x ^ 2 + 3x + 4, and find the tangent of the second point (x0, Y0) of the parabola

y=x^2+3x+4
y"=2x+3
The derivative at x0, i.e. tangent slope k = 2x0 + 3
Let the tangent point be (x0, x0 ^ 2 + 3x0 + 4)
x0^2+3x0+4/x0=2x0+3
X0 = 2 or - 2
So the tangent point is (2,14) or (- 2,2)
The tangent equation is y = 5x + 4 or y = - X

Tangent chord equation of circle
I mean, the general tangent chord equation is (x-a) 2 + (y-b) 2 = R2, not x2 + y2 = R2
The tangent chord equation (if even this is better) is the tangent chord equation rather than the common chord equation

No, the tangent chord equation of (x-a) 2 + (y-b) 2 = R2 is (x-a) (x0-a) + (Y-A) (y0-a) = R2
A B is the coordinates of the center of the circle, x0, Y0 is the coordinates of the point P passing through the outside of the circle. If P is on the circle, there is only one tangent line passing through the point P. when a = b = 0, it is the case of the origin
I wish you a happy study!

Given the point P (4,2) and the circular equation x ^ 2 + y ^ 2 = 10, make two tangent lines of the circle through P, and the tangent points are a and B. find the linear equation where the tangent point chord AB lies
2. Make any secant of a circle through point P, intersect the circle at C and D, and find the trajectory equation of point E in CD

The process x ^ 2, y ^ 2 = 10, make two tangent lines of the circle through P, and the tangent points are a and B

It is known that the equation of parabola P is x2 = 4Y, and the tangent of parabola is made through any point a on the straight line L: y = - 1. Let the tangent points be B and C respectively. (1) prove that △ ABC is a right triangle; (2) prove that the straight line BC passes through a fixed point, and find out the coordinates of the fixed point

(1) It is proved that: let a (m, - 1), B (x1, Y1), C (X2, Y2) ∵ the equation of parabolic p be x2 = 4Y, ∵ y ′ = 12x. ∵ Y1 + 1x1 − M = 12x1, ∵ 14x12 + 1 = 12x12-12mx1, ∵ x12-2mx1-4 = 0. Similarly, we can get that x22-2mx2-4 = 0, ∵ X1 + x2 = 2m, x1 · x2 = - 4. ∵ KAB · KAC = 12x1 · 12x2 = 14 · x1 · X & nbsp; (2) it is proved that the linear equation of BC is Y-Y1 = Y1 − y2x1 − X & nbsp; 2 (x-x1), and y-14x12 = 14 (x1 + x2) (x1-x2), that is, y = 12mx + 1. Obviously, when x = 0, y = 1, so the straight line BC passes through the fixed point (0, 1)

If M (2,1) is a point on the parabola x2 = 2PY, then the tangent equation of the parabola with m as the tangent point is______ ．

∵ point m (2,1) is a point on the parabola x2 = 2PY, ∵ P = 2, ∵ when the parabola equation is y = 14x2, ∵ y ′ = 12x, x = 2, y ′ = 1, ∵ the tangent equation of the parabola with point m as tangent point is Y-1 = X-2, that is, x-y-1 = 0

It is known that the parabolic equation x ^ 2 = 4Y passes through the point P (T, - 4) to make two tangent lines PA, Pb of the parabola, and the tangent points are a and B.10 respectively
It is known that the parabolic equation x ^ 2 = 4Y passes through the point P (T, - 4) as two tangent lines PA and Pb of the parabola, and the tangent points are a and B respectively
1) Verification: the straight line AB passes the fixed point (0,4);

We know the parabolic equation x & # 178; = 4Y, passing through the point P (T, - 4) as two tangent lines PA and Pb of the parabola, and the tangent points are a and B respectively. We prove that the straight line AB passes through the fixed point (0,4). Let the tangent equation of p be y = K (x-t) - 4, and substitute it into the parabolic equation to get X & # 178; - 4 [K (x-t) - 4] = x & # 178; - 4kx + 4kt + 16 = 0, and make its discriminant △ = 16

It is known that the parabolic equation x2 = 4Y, passing through the point (T, - 4) makes two tangent lines PA and Pb of the parabola, and the tangent points are a and B respectively
(1) Verify that the line AB passes through the fixed point (0,4);
(2) Find the minimum area of △ OAB (o is the coordinate origin)
(I) let the tangent point be a (x1, Y1), B (X2, Y2), and y '= X,
Then the equation of tangent PA is Y-Y1 = X1 (x-x1), that is y = X-Y1,
The equation of tangent Pb is: Y-Y2 = (x-x2), that is, y = x-y2,
From (T, - 4) is the intersection of PA and Pb, we can see that: - 4 = x1t-y1, - 4 = x2t-y2,
The linear equation passing through a and B is - 4 = tx-y,
That is, tx-y + 4 = 0, so the line AB: tx-y + 4 = 0 passes through the fixed point (0,4)
(II) from X 2-2tx-16 = 0
Then X1 + x2 = 2T, x1x2 = - 16,
Because s △ OAB = × 4 ×| x1-x2 | = 2 = 2 ≥ 16, if and only if t = 0, s minimum = 16
Just search the answer on the Internet, where y '= x is why?

It is to find the derivative of the parabolic equation x ^ 2 = 4Y, that is to say, to find the slope. After finding the slope, it is brought into the point oblique tangent equation PA and Pb

It is known that parabola C: x ^ 2 = 4Y, straight line L: y = - 1, PA and Pb are two tangent lines of curve C, and the tangent points are a and B respectively. If P is on L, PA ⊥ Pb is proved

Let's talk about the following ideas: let P (m, - 1), and then set any point of the parabola (n, n ＾ 2 / 4). In this way, we can find the tangent equation of n points, only including xyn, passing through P points, substituting P into the tangent equation, including Mn, and finding out two relations (using one to represent the other). There should be two kinds, that is, the representation of AB point about P points, and then verify that the vector product of PA and Pb is 0

What are the focal coordinates and the Quasilinear equation of the parabola y = 2px?