The value range of the slope of a straight line passing through point (0,2) and hyperbola x ^ 2 / 9-y ^ 2 / 16 = 1 with only one common point Online, etc

The value range of the slope of a straight line passing through point (0,2) and hyperbola x ^ 2 / 9-y ^ 2 / 16 = 1 with only one common point Online, etc

Let the line be y = KX + A, because it passes through (0,2) point, we can get that a = 2Y = KX + 2 and x ^ 2 / 9-y ^ 2 / 16 = 1 have and only have one common point, that is, the equation system {x ^ 2 / 9-y ^ 2 / 16 = 1; y = KX + 2} has only one solution. Substituting y = KX + 2 into x ^ 2 / 9-y ^ 2 / 16 = 1, we can get (16-9k ^ 2) x ^ 2-18kx-180 = 0 when 16-9k ^ 2 = 0, the equation

If the line y = - x + 2 √ 6 and the image of hyperbola y = K / X have and only have one common point P?

Suppose that the common point is x0, Y0, then there is - x0 + 2 √ 6 = K / x0, i.e. - x0 square + 2 √ 6x0-k = 0. Because there is only one common point, the equation has only one solution, which is solved by b square - 4ac = 0, i.e. (2 √ 6) square - 4 (- 1) (- K) = 0, and the solution is k = 6

If the inclination angle of a straight line is 75 degrees, then the slope of the straight line is

Because the slope of a straight line is the tangent of the inclination angle of a straight line, so
K = tan75 ° = Tan (30 ° + 45 °) = (tan30 ° + tan45 °) / (1-tan30 ° tan45 °) = 2 + radical 3
(Tan 30 ° = (radical 3) / 3, Tan 45 ° = 1)

If the straight line passes through the point P (2,3) and the inclination angle is half π, the slope of the straight line can be calculated

The equation at 90 ° is perpendicular to the x-axis
So x = 2
The slope does not exist

The standard equation for finding a circle whose center is on a straight line 2x-y = 3 and tangent to two coordinate axes

Let the circle coordinate be (x, x) or (- x, - x) and substitute the line 2x-y = 3 to get x = 3, or x = - 3
So the standard equation of circle is: (x-3) ^ 2 + (Y-3) ^ 2 = 9 or (x + 3) ^ 2 + (y + 3) ^ 2 = 9

Find the circle equation with the origin as the center and tangent to the line 2x + 3Y + 7 = 0

r=d=7/√13
∴x²＋y²=49/13

Find the equation of a circle suitable for the following conditions: 1. The center of the circle is (3,4) and passes through the origin; 2. The center of the circle is at the origin and tangent to the line 4x-3y-15 = 0
Find the equation of a circle suitable for the following conditions: (1) the center of the circle is (3,4) and passes through the origin; (2) the center of the circle is at the origin and tangent to the line 4x-3y-15 = 0

Solution
The radius is five
So the standard equation for a circle is
(x-3)^2+(y-4)^2=25
The center of the circle is at the origin
So the distance from the origin to the line is
/15/ /5=3
So the radius is three
So the equation for a circle is
x^2+y^2=9

Let a straight line pass through a point (3,5) and its inclination angle is 45 degrees greater than that of the straight line 3x-2y + 7 = 0, then the linear equation is
Through the point (2, - 1), the linear equation of inclination angle is 1 / 2 of the inclination angle of the line 4x + 3y-1 = 0——

The inclination angle of the straight line L is 60 degrees larger than that of the straight line y = √ 3x, and the longitudinal intercept of the straight line L is 3

The slope of the line y = √ 3x is √ 3, and the inclination angle is 60 degrees
The inclination angle of L is 60 ° + 60 ° = 120 °
That is, the slope of the straight line L is k = tan120 ° = - 3
The longitudinal intercept of a straight line is 3
The equation of a straight line is y = - √ 3x + 3

If the inclination angle of a (- 1, - 3) is equal to 2 times of the inclination angle of the straight line y = 3x, can we use the point oblique formula

y＝3x
tana=3
Then k = tan2a = 2tana / (1-tan & # 178; a) = - 3 / 4
So y + 3 = - 3 / 4 * (x + 1)