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In the plane rectangular coordinate system, after the straight line y = 3x is shifted to the left by one unit length, the linear relationship is
After the straight line y = 3x is shifted to the left by one unit length, its linear relationship is
Y = 3 (x + 1), arranged as y = 3x + 3
After the straight line y = 3x is shifted to the right by one unit length, its linear relationship is
Y = 3 (x-1), sorted as y = 3x-3
In the plane rectangular coordinate system, the straight line y = - 2x + 1 is shifted to the left by 4 length units, and the analytic formula of the straight line is obtained It's minus 2x
In translation, left plus right minus, that is, left shift is solved by substituting (x + offset) into the original formula
The original formula becomes: y = - 2 (x + 4) + 1
The result is: y = - 2x-7
In the plane rectangular coordinate system, after the straight line y = x is shifted to the left by one unit length, its analytical formula is______ .
According to the principle of "left plus right minus", in the plane rectangular coordinate system, after the straight line y = x is shifted to the left by one unit length, the,
The analytical expression of the straight line is y = x + 1
So the answer is y = x + 1
As shown in the figure, in the plane rectangular coordinate system xoy, two lines a and B parallel to the X and Y axes intersect at point a (3, 4). Connect OA. If there is a point P on the line a, make △ AOP an isosceles triangle. Then the coordinates of all points P satisfying the conditions are______ .
∵A（3，4）
∴OB=3，AB=4
∴0A=
OB2+AB2=5
If OA is an isosceles triangle with one waist, then the coordinates of point P are (8,4) (- 2,4) (- 3,4);
When OA is the bottom edge,
∵A（3，4），
The analytical formula of straight line OA is y = 4
3x，
The analytic formula of the straight line passing through the midpoint of line OA and perpendicular to the line OA is: y = - 3
4x+25
8，
The coordinate of point P is (- 7
6，4）．
Therefore, fill in (8, 4) or (- 2, 4) or (- 3, 4) or (- 7)
6，4）．
As shown in the figure, in the plane rectangular coordinate system xoy, point F is an ellipse C: x2 a2+y2 If the quadrilateral ofmn is a diamond, then the eccentricity of ellipse C is______ .
According to the meaning of the title, the side length of the diamond is C, and the abscissa of the N point is C according to the symmetry of the ellipse
Since on = C, C2
4 + y2 = C2
Three
2c, then NF=
(
3C
2)2+(c
2+c)2=
3C
From the symmetry of the ellipse, we know that the distance from point n to the right focus is C, and from the definition of ellipse, 2A = C+
3C, so e = 2
1+
3=
3-1
So the answer is:
3-1
In the plane rectangular coordinate system xoy, the line x + Y - √ passing through the ellipse M: the square of the fraction of a plus the square of the fraction y of B is equal to the line x + Y - √ passing through the right focus of 1 = 0 crosses m at a, B and P is the midpoint of AB, and the slope of OP is 1 / 2 Q: find the equation of M
(I) put the right focus (C, 0) into the line x + y-
Three
=0 leads to C + 0-
Three
=0, C is obtained=
Three
.
Let a (x1, Y1), B (X2, Y2), the midpoint P (x0, Y0) of segment ab,
be
x 21
A2
+
y 21
B2
=1,
x 22
A2
+
y 22
B2
=1, minus
x 21
-
x 22
A2
+
y 21
-
y 22
B2
=0,
Qi
x1+x2
A2
+
y1+y2
B2
X
y1-y2
x1-x2
=0,
Qi
2x0
A2
+
2y0
B2
X, Kop) = 0=
One
Two
=
Y0
X0
,
Qi
One
A2
-
One
2B2
=0, that is, A2 = 2B2
United
a2=2b2
a2=b2+c2
C=
Three
It is found that
b2=3
a2=6
,
The equation of M is
X2
Six
+
Y2
Three
=1．
In the plane rectangular coordinate system xoy, it is known that the left focus of ellipse C: (x^2/a^2) + (y^2/b^2) =1 is F: (-1,0), and point P (0,1) is on C (1) Solving the equation of elliptic C (2) Let l be tangent to ellipse C and parabola C: y = 4x at the same time, and find the equation of line L
Because the left focus of ellipse C1 is F1 (- 1,0)
So c=1
Substituting point P (0,1) into ellipse
（x²/a²)+(y²/b²)=1
The result is: 1 / B 2 = 1
∴b=1
So a? =b? +c? =2
So the equation of elliptic C is 1 / 2 + y 2 = 1
The slope of the line L obviously exists
Let the equation of the line l be y = KX + M
½x²+y²=1
y=kx+m
By eliminating y and finishing, we get (1 + 2K? X 2) + 4kmx + 2m? - 2 = 0
Because the line L is tangent to the ellipse C
Δ=16k²m²-4(1+2k²)(2m²-2)=0
Finishing: 2K 2 - M 2 + 1 = 0
y²=4x
y=kx+m
By eliminating y and finishing, K? X? 2 + (2km-4) x + m? 2 = 0
Because the line L is tangent to the parabola C2
So △ = (2km-4) 2 - 4K M2 = 0
Finishing: km = 1 ②
According to (1) and (2)
k=√2/2 m=√2
or
k-√2/2 m=-√2
So the equation of the line L is:
y=√2/2x+√2
perhaps
y=-√2/2x-√2
On the ellipse x ^ 2 + (y ^ 2) / 2 = 1, the point P (- radical 2 / 2, - 1) crosses the ellipse and a, B with the slope of - radical 2,
Ellipse x ^ 2 + (y ^ 2) / 2 = 1
Focus coordinates (0,1) (0, - 1)
P (- root 2 / 2, - 1), the line passing through the focus and slope of - root 2 does not exist
What's wrong with this question?
The eccentricity of the elliptic equation is the root of two-thirds sign three. There are two intersections A and B between the straight line passing through the right focus F and the ellipse. If the vector AF = 3, the vector FB, calculate the slope K
K= ±2 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
The point a (0,1) is known; the straight line L with slope k intersects the circle C: (X-2) ^ 2 + (Y-3) ^ 2 = 1 at two different points m and N. 1) find the value range of real number K 2) 3) if O is the origin of coordinate, and the vector om * vector on = 12. Find the value of K. note: l but point a should be handed over in the afternoon,
1) If the straight line is Y-1 = KX, that is, y = KX + 1, it is brought into the circle equation to obtain (1 + K ^ 2) x ^ 2 - (4K + 4) x + 7 = 0. If there are two different points, then x must be different 2 values, that is, △ = (4K + 4) ^ 2-4 * 7 (1 + K ^ 2) > 0
Simplified 3K ^ 2-8k + 3
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