The slope of a straight line can be calculated when the inclination angle of the straight line is known (1)α=2 (2)α=89°

The slope of a straight line can be calculated when the inclination angle of the straight line is known (1)α=2 (2)α=89°

The slope of the straight line k = Tan α;
（1）k=tan α =tan 2=α =π*2/180=π/90；
（2）k=tan α =tan （89°）=sin(89°)/cos(89°)=1/[1-(π/180)^2]^0.5；

If Sina = 3 / 5, calculate the slope of the line

That is, 2: (1-x) = 4:5
4(1-x)=2×5=10
1-x=10÷4=2.5
x=1-2.5
x=-1.5
5²-x²=12²-(13-x)²
25-x²=144-169+26x-x²
26x=50
x=25/13

(1) given two straight lines X + 2y-4 = 0 and 3x + 6y-2 = 0, find the distance between two straight lines
(2) given that the intersection point of two straight lines X + 2y-4 = 0 and y = KX + 1 is in the fourth quadrant, the value range of K is obtained

3x 6y-2 = 0, that is, X 2y-2 / 3 = 0
Distance = | - 4 2 / 3 | / √ 5 = 2 √ 5 / 3
Simultaneous solution of X, y
x>0,y

If two lines 3x + 2y-3 = 0 and 6x + my + 1 = 0 are parallel to each other, then the distance between them is equal to?

If two lines are parallel, it means that the slope is the same, so 3:2 = 6: m  M = 4, and point a (1,0) is the point on the line 3x + 2y-3 = 0, then the distance between two lines is the distance from point a to line 6x + 4Y + 1 = 0 = (6 × 1 + 4 × 0 + 1) / (√ 6 × 6 + 4 × 4) = 7 / √ 52 = 7 / 2 √ 13

F1 and F2 are the two focal points of the ellipse x2 / 25 + Y2 / 16 = 1. If | ab | = 5, then | AF1 | + | BF1=|

(25 + 12 / 25 + Y22 / 16 = 1A = 5,2a = 5,2a = 10 [[[AF1 || [af2 || [BF1 || [BF1 | [BF1 [[[BF1 | [[[BF1 [[[BF1 [[[BF1 [[BF1 [[[BF1]]]] [[BF1 [[BF1 [[BF1 [[[BF1 [[[[BF1 [[[[[[[BF1]]]] [[[[BF1 [BF1 [BF1]]] [[[[[BF1 [[[BF1]]]]]]]]]] [[[[[[[| = 2A + 2A = 10 + 10 = 20, so | AF1 | + | BF1 | = 20-5 = 15

It is known that F1 and F2 are the left and right focal points of the ellipse C: x2a2 + y2b2 = 1 (a > b > 0) respectively. The straight line passing through F1 and perpendicular to X axis intersects ellipse C at two points a and B. If △ abf2 is a right triangle, then eccentricity e of ellipse C is ()
A. 2-1B. 3-1C. 22D. 33

In the ellipse C: x2a2 + y2b2 = 1 (a > b > 0), △ abf2 is a right triangle. According to the symmetry of the ellipse, we get | AF1 | = | F1F2 |, that is, B2A = 2C; | A2 − C2A = 2c, that is, 1e-e-2 = 0; we get e = 2-1, or E = - 2-1 (rounding); we get the heart rate of ellipse C E = 2-1

If F2, which passes through the right focus of the ellipse X225 + y216 = 1, makes a straight line perpendicular to the X axis and intersects with the ellipse at two points a and B, and F1 is the left focus of the ellipse, then the perimeter of △ af1b is ()
A. 10B. 20C. 30D. 40

∵ F1, F2 are the two focal points of the ellipse X225 + y216 = 1, | AF1 | + | af2 | = 10, | BF1 | + | BF2 | = 10, | △ af1b's perimeter is | ab | + | af2 | + | BF2 | = | AF1 | + | af2 | + | BF1 | + | BF2 | = 10 + 10 = 20

Through the left focus F1 of the ellipse x ^ 2 / 16 + y ^ 2 / 9 = 1, make the intersection ellipse perpendicular to the major axis at two points a and B, and F2 is the right focus, then | af2 | =?
Expressed as a false fraction

c=√(a^2-b^2)=√(16-9)=√7
Left focus F1 (- √ 7,0)
Substituting x = - √ 7 into x ^ 2 / 16 + y ^ 2 / 9 = 1,7 / 16 + y ^ 2 / 9 = 1, y = ± 9 / 4, i.e. a and B coordinates (- √ 7, ± 9 / 4)
|AF2| = √[(xF2-XA)^2+(yF2-yA)^2] = √[(√7+√7)^2+(0±9/4)^2] = √(529/16) = 23/4

As shown in the figure, the elliptic equation is x216 + y2b2 = 1 (4 & gt; B & gt; P is the moving point on the ellipse, F1 and F2 are the two focal points of the ellipse. When p is not on the x-axis, f1m is the vertical line of the bisector of the outer angle of ∠ f1pf2 through F1, and the perpendicular foot is m. when p is on the x-axis, M is defined to coincide with P (1) (2) given o (0,0), e (2,1), try to explore whether there is such a point Q: q is the integral point of the interior of the trajectory t (the point whose abscissa and ordinate are integers in the plane is called integral point), and the area of △ OEQ is s △ OEQ = 2? If it exists, find out the coordinate of point Q. if it does not exist, explain the reason

(1) When the point P is not on the x-axis, the extension line of extension f1m and F2P intersects at the point n, connecting OM, ∵ {NPM =} mpf1, ∵ {NMP =} PMF1 ≌ PNM ≌ pf1m ≌ m is the midpoint of the line NF1, | PN | = | Pf1 | (2 points) | om | = 12 | f2n | = 12 (| F2P | + | PN |) = 12 (| F2P | + | Pf1 |) | point P is in the

As shown in the figure, it is known that the eccentricity e of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is 1 / 2, the left and right focal points are F1 (- 1,0) and F2 (1,0), and the upper fixed point is a
Point Q (- 4,0) is a point outside the ellipse
(1) The equation of the line QA of the spherical ellipse C
The top vertex is a

Ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0)
The left and right focus were F1 (- 1,0), F2 (1,0)
C = 1, e = C / a = 1 / 2, a = 2
∴b²=a²-c²=4-1=3
｛ ellipse C: X & ｛ 178 / 4 + Y & ｛ 178 / 3 = 1
Q (- 4,0), fixed point a, who is it?