Find the minimum area of the tangent of the circle x ^ 2, y ^ 2 = 1 and the triangle surrounded by two coordinate axes, and find the equation of the tangent when the minimum value is 0 Let the tangent point (a, b), then let the equation AX + by = 1. Why can we set this tangent equation

Find the minimum area of the tangent of the circle x ^ 2, y ^ 2 = 1 and the triangle surrounded by two coordinate axes, and find the equation of the tangent when the minimum value is 0 Let the tangent point (a, b), then let the equation AX + by = 1. Why can we set this tangent equation

The tangent equation is obtained by setting the tangent point, and the intersection coordinates of the tangent and the coordinate axis are obtained respectively to express the area of the triangle surrounded by the tangent and the two coordinate axes, and then the minimum area can be obtained by using the basic inequality
Let the tangent point coordinates be (x0, Y0), because the slope of the tangent equation is perpendicular to the line passing through the tangent point radius, and the slope of the line passing through the tangent point radius is y0x0, then the slope of the tangent equation is - x0y0, so the tangent equation is y-y0 = - x0y0 (x-x0), because the tangent point is on the circle, so X02 + Y02 = 1, and the tangent equation is reduced to x0x + y0y = 1,
The intersection coordinates of the tangent and the two axes are (1x0,0), (0,1y0), respectively,
Therefore, the area of the triangle bounded by the tangent and the two coordinate axes is 12x0y0, and X02 + Y02 = 1,
So 12x0y0 ≥ 1x02 + Y02 = 1, that is, the minimum area of the triangle surrounded by the tangent and the two coordinate axes is 1
So the answer is 1

Find the area enclosed by the parabola y ^ 2 = X-1 and its tangent at points (2,1), (2, - 1)

The tangent slope at (2,1) is 1 / 2
The equation is Y - 1 = 1 / 2 (x - 2)
The intersection of this line and X axis is (0,0)
Figure area = triangle area - area between parabola and x = 2
Parabola area (half) = integral (YDX) = integral (sqrt (x) DX) (0 - > 1) = 2 / 3
Enclosed area = 2 - 4 / 3 = 2 / 3

If the area of the triangle formed by the tangent of the parabola y = x ^ 2 at the point (a, a ^ 2) and the two coordinates is 16, find a

y'=2x
y'=2a
Tangent Y-A & #178; = 2A (x-a)
y=2ax-a²
x=0
y=-a²
y=0
x=a/2
a²|a/2|*(1/2)=16
a=±4

Find a point on the parabola y = x ^ 2 so that the tangent of the point and the straight line y = O, x = 8 form a triangle with the largest area

Y '= 2x, let the tangent point be m (T, T & # 178;), then the tangent slope k = 2T, then the tangent equation is:
If 2tx-y-t & # 178; = 0, the intersection with the straight line y = 0 is Q (T / 2,0), and the intersection with the straight line x = 8 is p (8,16t-t & # 178;), then the area of triangle is:
S = (1 / 2) × [8 - (T / 2)] × (16t-t & # 178;), where 0

Given that the line L passes through the point P (- 1,0) and intersects with the parabola y ^ 2 = 2x at two points a and B, the trajectory equation of the midpoint m of the line AB is obtained

The straight line with slope 1 intersects with x ^ 2 = 2Y of parabola and two points a and B, so the trajectory equation of the midpoint of chord AB is

Let a (x1, Y1) B (X2, Y2), AB midpoint (x0, Y0). From the great theorem, we get X1 + x2 = 2, so x0 = 1, Y0 = 1 + B. from the discriminant greater than zero, we get b > - 1 / 2, so Y0 > 1 / 2
In conclusion, the trajectory equation is x = 1 (Y > 1 / 2)

Given that P (a, b) can make two tangent lines PA, Pb of parabola y = x ^ 2, the tangent point is a, B, if the angle APB = 90, find the trajectory of point P

Two tangent lines PA, Pb of parabola y = x ^ 2 can be made through point P (a, b)
Let a (x1, X & # 178; 1), B (X2, X & # 178; 2)
Derivation y '= 2x
The slope of PA is K1 = 2x1
The slope of Pb is K2 = 2x2
∵ angle APB = 90 & # 186;,
∴k1k2=-1
That is, x1x2 = - 1 / 4
PA equation Y-X & # 178; 1 = 2x1 (x-x1)
Pb equation Y-X & # 178; 1 = 2x2 (x-x2)

Two tangent lines PA of parabola through point P (3 / 2, - 1) are perpendicular to Pb, then a =?

The straight line passing through the point P is y = k [x - (3 / 2)] - 1, which is connected with the known parabola. By eliminating y, we obtain a quadratic equation with the letter K and with respect to X. then the equation can also be regarded as a quadratic equation with respect to K. if the solution K1 and K2 satisfy that the product of the two is - 1 [perpendicular], then we can calculate the value of X and solve the problem

The focal point of y = X2 is f, the moving point P moves on the straight line x-y-2 = 0, passing through the point P to make two tangent lines PA and Pb of the parabola, which are tangent to a and B respectively
1) Finding the trajectory equation of the center of gravity g of Δ APB
2) It is proved that: ∠ PFA = ∠ PFB

Given that the fixed point (0,3), the ellipse x squared / 9 + y = 1, and the point m (x, y) is a moving point on the ellipse, the maximum value of | Ma | is obtained
Hope to solve this problem with graphics!

With parameters
x=3cosa,y=sina
|MA|=√[(3cosa)^2+(3-sina)^2]
=√[9(cosa)^2+(sina)^2-6sina+9]
=√[18-8(sina)^2-6sina]
=√[-8(sina-3/8)^2+18+9/8]
When Sina = 3 / 8
|The maximum value of Ma | is √ (18 + 9 / 8) = √ 153 / 8