Parabola y2 = 4x, slope - 2 straight line intersection parabola at a, B two points, find a, B two points distance
Let a (x1, Y1) B (X2, Y2) linear equation y = - 2x + B Y1 ^ 2 = 4x1 (1) Y2 ^ 2 = 4x2 (2) as a difference, the deformation is: (y1-y2) / (x1-x2) = 4 / (Y1 + Y2) = - 2, Y1 + y2 = - 2, y = - 2x + B, y ^ 2 = 4x together, we get: 4x ^ 2-8x + B ^ 2 = 0, X1 + x2 = 2, Y1 = - 2x1 + B, y2 = - 2x2 + B
RELATED INFORMATIONS
- 1. It is known that the parabolic equation y2 = 4x, the line L passing through the fixed point P (- 2,1) and the slope k intersect the parabolic equation y2 = 4x at two different points. The range of the slope k is obtained
- 2. Given that the parabola y ^ 2 = 4x, the points P (1,2), a (x1, Y1), B (X2, Y2) are on the parabola, when the slopes of PA and Pb exist and the inclination angles complement each other, we can find the solution Find the value of Y1 + Y2 and the slope of line ab
- 3. Through the point m (2,0), make a straight line L with slope 1, intersect the parabola y * 2 = 4x at points a and B, and find the length of ab
- 4. It is known that the equation of parabola is y ^ 2 = 4x, the line L passes through the fixed point P (- 2,1), and the slope is K. when the value of K is, there is a common point between the line L and the parabola It is known that the equation of parabola is y ^ 2 = 4x, the straight line L passes through the fixed point P (- 2,1), and the slope is K. when the value of K is, the straight line L and parabola have one common point; there are two common points, but there is no common point
- 5. In the triangle ABO, vector OA = vector a, vector ob = vector B, M is the midpoint of ob, n is the midpoint of AB, P is on, the intersection of am, then what is AP equal to? Options: A 2 / 3 vector A-1 / 3 vector b B - 2 / 3 vector a + 1 / 3 vector b C 1 / 3 vector A-2 / 3 vector b D - 1 / 3 vector a + 2 / 3 vector b
- 6. In the triangle ABC, AC is 2, AB is 4, CB is 3, I is the heart, AI is XAB plus YAC, x plus y is what
- 7. I is the heart of △ ABC, AC = 2, CB = 3, Ba = 4, AI vector = XAB vector + YAC vector, then x + y =?
- 8. Given that | BC | = 2 in triangle ABC and point P satisfies AP vector = XAB vector + YAC vector, if x, Y > = 0, x + y = 1 / 2, then the length of the graph formed by P is? Don't understand what the last question is?
- 9. As shown in the figure, the triangle ABC is the inscribed triangle of circle O, I is the inner part of triangle ABC, the extension line of AI intersects BC at point E, and intersects circle O at point D Come on, now, now!
- 10. As shown in the figure, point I is the inner part of triangle ABC, the extension line of line AI intersects the circumcircle of triangle ABC at point D, and intersects the edge of BC at point E. prove that id = BD, BD square equals Da times EA The second question is more important!
- 11. When the straight line L passing through M (4.0) intersects the parabola C: y ^ 2 = 4x at two points a and B, 2am = MB, calculate the slope k of L RT,
- 12. It is known that the collimator of parabola C: y ^ 2 = 4x and X-axis intersect at point m, the straight line L with slope k of point m intersects with parabola C at two points ab 1 F is the focal point of parabola C. if the module am = 5 / 4, the module AF can find the value of K 2 whether there is such a K, so that there is always Q on the parabola C, and if QA vertical QB exists, ask for the value range of K
- 13. Given that the moving point P of a (- 3,8) B (7, - 4) satisfies the vector AP * vector BP = 0, then the trajectory equation of point P is
- 14. In △ ABC, a is an obtuse angle, Sina = 3 / 5, C = 5, B = 4
- 15. Given a = (1,2), B = (- 3,2), if Ka + B is parallel to a-3b, then the value of real number k is
- 16. Given a = (1,5, - 1), B = (- 2,3,5), if (KA + b) is parallel (a-3b), find K
- 17. Find the general equations of the following plane: passing through points (3,1, - 1) and (1, - 1,0), parallel to vector V = (- 1,0,2)
- 18. What is the linear equation passing through point (0,4) and parallel to vector V = (2,3)?
- 19. The point direction linear equation passing through point P (- 3,5) and parallel to vector V (5,2) is
- 20. How to use parallel vector (1,2) to get linear equation?