Given that the coordinates of a and B are (0, - 5) and (0,5) respectively, and the product of the slope of Ma and MB is λ, the trajectory equation of M is solved and the trajectory shape of M is judged~ Detailed process It's OK to have an answer~

Given that the coordinates of a and B are (0, - 5) and (0,5) respectively, and the product of the slope of Ma and MB is λ, the trajectory equation of M is solved and the trajectory shape of M is judged~ Detailed process It's OK to have an answer~

Let m (x, y) x ≠ 0
MA：k1=(y+5)/x
MB：k2=(y-5)/x
The product of slopes is λ
k1*k2=(y+5)(y-5)/x^2=λ
Simplification - λ x ^ 2 + y ^ 2 = 25
The trajectory equation of M: - λ x ^ 2 + y ^ 2 = 25 (x ≠ 0)
① When λ∈ (- ∝, - 1), the,
The trajectory is ellipse (except two points), and the focus is on the Y axis;
② When λ = - 1,
The trajectory is a circle (except two points), x ^ 2 + y ^ 2 = 25;
③ When λ∈ (- 1,0), the,
The trajectory is ellipse (except two points), and the focus is on the X axis;
④ When λ = 0,
The trajectory is two parallel straight lines (except two points), y = 5 or y = - 5;
① When λ ∈ (0, + ∝),
The trajectory is hyperbolic (except two points), and the focus is on the Y axis

The moving point m and two fixed points a (- 1,0)'b (1.0) form a triangle mAb, and a straight line MA.MB The product of the slopes of C is 4

Let the coordinates of point m be (x, y)
The slope of AM = Y / (x + 1)
The slope of BM = Y / (x-1)
The multiplication of the two is: Y & # 178; / [x & # 178; - 1] = 4
The deformations are as follows
x²-y²/4=1

As shown in the figure, M is a fixed point on the parabola y2 = x, the moving chords me and MF intersect with the X axis at different points a and B, and | Ma | = | MB |. It is proved that the slope of the straight line EF is a fixed value

Let K, the slope of the line me be K (k > 0), then the slope of the line MF is - K, and the equation of the line me is y-y0 = K (x-y02). From y − Y0 = K (x − Y02) y2 = x, we can get ky2-y + Y0 (1-ky0) = 0. Then y0ye = Y0 (1 − ky0) k, so ye = 1 − ky0k. Similarly, we can get YF = 1 + ky0 − K, ｜ KEF = ye − yfxe − XF = ye − yfye2 − yf2 = 1ye + YF = − 12y0 (fixed value)

As shown in the figure, AB is the moving chord on the parabola y = X2, and | ab | = a (a is constant and a ≥ 1). Find the shortest distance between the midpoint m of the chord AB and the X axis

Let the ordinates of a, m and B be Y1, Y2 and Y3 respectively. As shown in the figure, the projections of a, m and B on the parabola are a ', M' and B 'respectively. F is the focus of the parabola, connecting AA', mm ', BB', AF and BF