Given that the coordinates of the midpoint P of the chord ab of the ellipse X & # 178 / 6 + X & # 178 / 5 = 1 are (2, - 1), then the equation of the straight line AB is

Given that the coordinates of the midpoint P of the chord ab of the ellipse X & # 178 / 6 + X & # 178 / 5 = 1 are (2, - 1), then the equation of the straight line AB is

From the image, it is impossible to be perpendicular to the x-axis, so let the linear equation be y + 1 = K (X-2)
(5+6k^2)x^2-12k(2k+1)x+6(2k+1)^2-30=0
And X1 + x2 = 12K (2k + 1) / (5 + 6K ^ 2) = 4, the solution is k = 5 / 3
So the linear equation is y = 5 / 3x-13 / 3
It is known that, as shown in the figure, in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o
It is proved that in the plane ABC (? ∵ + ABC) = 12, ABC (? ∵ + ABC) = 12
The ellipse with the center of the linear equation at the origin passes through the point a (2,3) and the point F (2,0) is the right focus. The standard equation x & # 178 / 16 + Y & # 178 / 12 = 1
Ask if there is a line L parallel to OA
What I want to ask is how to set the linear equation. Why is it y = 3 / 2x + T? I know that the line is parallel to OA, so k = 3 / 2. But what's that t? Does it change like this (Y-T) = KX? It's not right. It doesn't intersect with y axis
Well, the line L parallel to OA has the same slope, but there is only one constant difference between the parallel lines. Therefore, the equation can be set as y = 3 / 2x + t
T is just a constant, just like B in y = KX + B is the same. The straight line should intersect the y-axis, because the straight line equation of OA is y = 3 / 2x, passing through the origin, the straight line is parallel to OA, so it must intersect the y-axis, t is the intercept on the y-axis.
It is known that: in △ ABC, the bisectors BD and CE of ∠ ABC and ∠ ACB intersect at O, ∠ ABC = 40 ° and ∠ ACB = 80 ° to find the degree of ∠ BOC
The bisectors BD and CE of ∵ - ABC and ∵ ACB intersect at point O, ∵ ABC = 40 °, ACB = 80 °, DBC = 12 ∵ ABC = 20 °, ECB = 12 ∵ ACB = 40 ° and ∵ BOC = 180 ° - DBC - ∵ ECB = 180 ° - 20 ° - 40 ° = 120 °. Answer: ∵ BOC = 120 °
The focus of the ellipse C: X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1 (a > 0, b > 0) is F1 (0,3)
Then: for the point on the ellipse C, the area of △ mof1 is 3 / 2, find the equation of the ellipse C. according to the specific situation of the answer, add 5-50 extra reward points, hard work!
The original problem should be: given the ellipse C with the center at the origin: X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a ﹥ 0, B ﹥ 0), a focus is F1 (0,3), m (x, 4) (x ﹥ 0) is a point on the ellipse C, the area of △ mof1 is 3 / 2, and the equation of ellipse C is solved.
According to your conditions, you can only know that the focus of the ellipse is on the y-axis, C = 3, the abscissa of point m is 1 or - 1, and nothing else can be determined
Focus F1 (0,3)
A point m (x, 4) on an ellipse
S△MOF1=3/2
X>0
According to the triangle area formula, 4 * | x | / 2 - (4-3) * | x | / 2 = 3 / 2
We get x = 1,,
That is to say, m (1,4) is on the ellipse and substituted into the equation
1/a^2+16/b^2=1
C=3
b^2-a^2=9
So a ^ 2 = 9, B ^ 2 = 18
So the elliptic equation is
X ^... Unfold
Focus F1 (0,3)
A point m (x, 4) on an ellipse
S△MOF1=3/2
X>0
There is a formula of | / 2 * | - 2 * | - 2 * | - 2 * | - 2 * | - 2 * | - 2 * | - 2 * | - 2 * | - 2 * | - 2 * | - 2 * | - 2 * | - 2 *
We get x = 1,,
That is to say, m (1,4) is on the ellipse and substituted into the equation
1/a^2+16/b^2=1
C=3
b^2-a^2=9
So a ^ 2 = 9, B ^ 2 = 18
So the elliptic equation is
X ^ 2 / 9 + y ^ 2 / 18 = 1_ Thank you. I don't know which is the best answer
It is known that: in △ ABC, the bisectors BD and CE of ∠ ABC and ∠ ACB intersect at O, ∠ ABC = 40 ° and ∠ ACB = 80 ° to find the degree of ∠ BOC
The bisectors BD and CE of ∵ - ABC and ∵ ACB intersect at point O, ∵ ABC = 40 °, ACB = 80 °, DBC = 12 ∵ ABC = 20 °, ECB = 12 ∵ ACB = 40 ° and ∵ BOC = 180 ° - DBC - ∵ ECB = 180 ° - 20 ° - 40 ° = 120 °. Answer: ∵ BOC = 120 °
Let the equation of line l be y = KX + B (where the value of K is independent of B), and the equation of circle m be X & # 178; + Y & # 178; - 2X-4 = 0. (1) if there are always two different intersection points between line L and circle M regardless of the value of K, the value range of B can be obtained; (2) when B = 1, l and circle intersect at two points a and B, the maximum and minimum value of AB can be obtained
(1) The standard equation of circle is: (X & # 178; - 1) + Y & # 178; = 5,
The center of the circle (1,0), on the x-axis, the radius r = √ 5, and the two intersections with the y-axis are (0,2), (0, - 2)
The intersection of the line y = KX + B (where the value of K is independent of B) and the Y axis is (0, b)
Because no matter what the value of K is, there are always two different intersections between the line L and the circle M
So the intersection of line and Y axis is between circle and Y axis, that is - 2
A quadratic equation about X can be obtained by eliminating y from the equation of line and circle. Because there are two intersections between line and circle, the discriminant is greater than zero. The discriminant contains k and B, which is a quadratic inequality about K. because no matter what the value of K is, it becomes a quadratic inequality about K
If the quadratic inequality of K is greater than zero, let the discriminant be less than zero. The discriminant only contains B, so the range of B can be solved.
The second question is still simultaneous equation, the focus has two, so we can solve the discriminant is greater than zero, and then determine the range of K, and then use the chord length formula to solve the maximum value, in the root formula is the restricted quadratic function... Expansion
A quadratic equation about X can be obtained by eliminating y from the equation of line and circle. Because there are two intersections between line and circle, the discriminant is greater than zero. The discriminant contains k and B, which is a quadratic inequality about K. because no matter what the value of K is, it becomes a quadratic inequality about K
If the quadratic inequality of K is greater than zero, let the discriminant be less than zero. The discriminant only contains B, so the range of B can be solved.
The second question is still a simultaneous equation with two focuses, so the discriminant is greater than zero, and then the range of K can be determined. Then the chord length formula is used to solve the maximum value. In the root formula is the problem of solving the range of quadratic function with restriction.
There's no time to edit the detailed answers. Just do it the way I said. Put it away
In △ ABC, BD and CE divide ∠ ABC and ∠ ACB equally, and BD and CE intersect at point O. if ∠ a = 50 degrees, then ∠ BOC = ()
105 degrees
It is known that the ellipse X & # 178; / 8 + Y & # 178; / 6 = 1 and the line L: y = KX + T tangent to the circle (x-1) &# 178; + Y & # 178; = 1 intersect the ellipse at M and n,
If a point C on the ellipse satisfies om vector + on vector = λ OC vector, find the value range of real number λ
The relationship between K and t can be obtained by the simultaneous circular and tangent equations, and the coordinates of M and N can be obtained by the simultaneous linear and elliptic equations. The coordinate expressions of OM and on (expressed by K and T) can be obtained by using Weida's theorem. In this way, the coordinates of C can also be expressed, that is, the intersection of the line where the vector OM and on are located and the ellipse, so that the length of OM + on and the length of OC are both
In the acute triangle ABC, the angle a = 40 degrees, the intersection of two high BD and CE with point O, then the degree of the angle BOC is-------
One hundred and forty