If there is a common point between the line L passing through point a (4,0) and the curve (X-2) + y = 1, what is the range of the slope of the line l?

Let the slope of the straight line l be K, then the equation of the straight line L is y-0 = K (x-4) ① because the linear equation L and the curve equation: (X-2) + y = 1 ② have a common point simultaneous equation ① ② there will be a new equation to solve the equation system. In the new equation, let △ ≥ 0, then the value range of the slope k can be obtained~

If there is only one common point between the straight line L passing through the fixed point (- 2,1) and the curve C: y = 3 - √ (4x-x ^ 2), the value range of the slope of the straight line L is,
It's the root

Y = 3 - √ (4x-x ^ 2), (Y-3) ^ 2 + (X-2) ^ 2 = 4 is a circle with a (2,3) as its center and 2 as its radius. Y = 3 - √ (4x-x ^ 2) is the lower half of the circle. When l: k = 0, l is tangent to the lower half of the circle, and there is only one common point. When k = 1 / 3, l intersects the lower half of the circle at the right end, and then it intersects the lower half of the circle

If there is a common point between the line L passing through point a (4,0) and the curve (X-2) ^ 2 + y ^ 2 = 1, then the value range of the slope of the line is?
It's filling in the blanks

Y = K (x-4) is substituted into the curve
(1+k^2)x^2+(8k^2-4)x+16k^2+3=0
(8k^2-4)^2-4(1+k^2)(16k^2+3)>=0
35k^2

If there is only one common point between the straight line passing through point (1, - 3) and hyperbola x ^ 2-y ^ 2 = 4, find the slope k of the straight line

If there is only one common point between the straight line passing through point (1, - 3) and hyperbola X & # 178; - Y & # 178; = 4, find the slope k of the straight line
Hyperbola X & # 178 / 4 + Y & # 178 / y = 1 is an equiaxed hyperbola, its a = b = 2, so its asymptote is y = ± X; so it is a straight line passing through point (1, - 3)
When the line is parallel to the two asymptotes, that is, when its slope k = ± 1, there must be only one intersection point with the hyperbola
In addition, two tangent lines of hyperbola can be made through (1, - 3). For this reason, let the equation of the line be y = K (x-1) - 3 = KX - (K + 3);
Substituting into hyperbolic equation, we get X & # 178; - [KX - (K + 3)] &# 178; - 4 = (1-k & # 178;) x & # 178; + 2K (K + 3) x - (K + 3) &# 178; - 4 = 0;
Because there is only one intersection, the discriminant is as follows
Δ=4k²(k+3)²+4(1-k²)[(k+3)²+4]=0
It is reduced to (K + 3) ² + 4 (1-k & #178;) = K & #178; + 6K + 9 + 4-4k & #178; = - 3K & #178; + 6K + 13 = 0, that is, 3K & #178; - 6k-13 = 0
So k = (6 ± √ 192) / 6 = (6 ± 8 √ 3) / 6 = 1 ± (4 / 3) √ 3
Conclusion: k = ± 1 or K = 1 ± (4 / 3) √ 3)
I hope it can help you

The focus of hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) is F1 and F2, the chord AB passes F1 and a and B are on the same branch
If | af2 | + | BF2 | = 2 | ab |, the perimeter of triangle abf2 is a, 4A B, 8A C, 12a D, which cannot be determined

|AF2|-|AF1|=2a
|BF2|-|BF1|=2a
So,
|AF2|+|BF2|=(2a+|AF1|)+(2a+|AF2|)
=4a+(|AF1|+|BF1|)
=4a+|AB|
So, 4A + | ab | = 2 | ab|
|AB|=4a
Perimeter of triangle abf2
=|AF2|+|BF2|+|AB|
=2|AB|+|AB|
=3|AB|
=12a
Choose C

The length of the chord AB obtained from the square section line y = x + 1 of hyperbola x-4 / Y

x^2-y^2/4=1
y=x+1
Lianlide: 4x ^ 2 - (x + 1) ^ 2 = 4
The reduction is: 5x ^ 2-2x-3 = 0
Two (x1-x2) ^ 2 = ⊿ / A ^ 2 = (2 ^ 2 + 4 * 5 * 3) / 5 ^ 2 = 64 / 25
According to the linear equation (y1-y2) ^ 2 = (x1-x2) ^ 2
So: ab = √ {2 (x1-x2) ^ 2} = 8 √ 2 / 5

The center is at the origin, a hyperbola with focus f (1,0), the ratio of the actual axis length to the imaginary axis length is m, and the standard equation of hyperbola is obtained

The center is at the origin, and a focus is f (1,0) so C = 12a / 2B = ma = BMA & sup2; + B & sup2; = C & sup2; = 1 so B & sup2; = 1 / (M & sup2; + 1) a & sup2; = B & sup2; M & sup2; = M & sup2; / (M & sup2; + 1) so (M & sup2; + 1) x & sup2 / / M & sup2; - (M & sup2; + 1) y & sup2; = 1

The center of the hyperbola is at the origin, the focus is on the x-axis, the eccentricity is 3, and the standard equation of the hyperbola is solved by (- 3,8)

E = C / a = 3C = 3AB & sup2; = C & sup2; - A & sup2; = 8A & sup2; the center is at the origin, the focus is on the x-axis, X & sup2; - Y & sup2; / 8A & sup2; = 1 through (- 3,8) 9 / a & sup2; - 64 / 8A & sup2; = 11 / A & sup2; = 1A & sup2; = 1x & sup2; - Y & sup2; = 8 = 1

Find a hyperbolic equation with the center at the origin, a focal point F (0,2), and the abscissa of the midpoint of the chord cut by the straight line y = x + 2 as 1

a. What is B

The center of hyperbola is at the origin, the focus is on the x-axis, the distance between the two collimators is 9 / 22, and it intersects with the straight line y = 1 / 3 (x-4). The abscissa of the middle point of the chord is - 2 / 3

Let the hyperbolic equation be X & sup2; / A & sup2; - Y & sup2; / B & sup2; = 1
Since the distance between the two collimators is 9 / 22, then 2A & sup2 / / C = 9 / 22, that is a & sup2; = 9C / 44
Simultaneous equations y = 1 / 3 (x-4) and X & sup2 / A & sup2; - Y & sup2 / B & sup2; = 1
The result is: X & sup2; (9b & sup2; - A & sup2;) + 8A & sup2; x-16a & sup2; - 9A & sup2; B & sup2; = 0
So X1 + x2 = - 8A & sup2; / (9b & sup2; - A & sup2;)
The abscissa of the midpoint is - 2 / 3, so X1 + x2 = - 4 / 3
So - 4 / 3 = - 8A & sup2; / (9b & sup2; - A & sup2;)
By sorting out the above formula and substituting C & sup2; - A & sup2; for B & sup2;, 7a & sup2; = 9 (C & sup2; - A & sup2;)
That is 16A & sup2; = 9C & sup2;
Since a & sup2; = 9C / 44, the solution is C = 4 / 11
So C & sup2; = 16 / 121
Then a & sup2; = 9 / 121
Then B & sup2; = C & sup2; - A & sup2; = 7 / 121
So the equation is X & sup2; / (9 / 121) - Y & sup2; / (7 / 121) = 1

If there is a common point between the line L passing through point a (4,0) and the curve (X-2) + y = 1, what is the range of the slope of the line l?