In △ ABC and △ def, ab = 4, BC = 5, AC = 8, de = 6, DF = 12 are known, then △ ABC is similar to △ def when EF = ()

In △ ABC and △ def, ab = 4, BC = 5, AC = 8, de = 6, DF = 12 are known, then △ ABC is similar to △ def when EF = ()

EF=（7.5）
4/6 =8/12 =2/3
5 / (2/3) = 7.5

The two ends of △ ABC are B (0,6) and C (0, - 6), and the product of the slopes of the straight lines on both sides is 4 / 9. It is very important to find the trajectory equation of vertex a

Let a be (x, y), the slope of line AB be K1, the slope of line AC be K2, K1 · K2 = 4 / 9, then [(y-6) / x] · [(y + 6) / x] = 4 / 9, and Y ^ 2 / 36-x ^ 2 / 81 = 1

The Quasilinear of parabola C: y square = 8x intersects with X axis at point P, a straight line passing through point P with positive slope k intersects with C at two points a and B, and F is the focus of C. if the absolute value of FA = the absolute value of 2fb, then =

The Quasilinear of parabola C: y ^ 2 = 8x ①: x = - 2 intersects with X axis at P (- 2,0),
A straight line with positive slope k passing through point P: y = K (x + 2) ② intersecting C at two points a (x1, *), B (X2, *),
If the focus of C is f, then | FA | = X1 + 2, | FB | = x2 + 2,
Substituting 2 into 1, K ^ 2x ^ 2 + (4K ^ 2-8) x + 4K ^ 2 = 0,
X = [4-2k ^ 2 Soil 4 √ (1-k ^ 2)] / K ^ 2,
From | FA | = 2 | FB |, we get X1 + 2 = 2 (x2 + 2), X1 = 2x2 + 2,
[4-2k^2+4√（1-k^2)]/k^2=2[4-2k^2-4√（1-k^2)]/k^2+2,
Multiply both sides by K ^ 2 to get 4-2k ^ 2 + 4 √ (1-k ^ 2) = 8-2k ^ 2-8 √ (1-k ^ 2),
It is reduced to √ (1-k ^ 2) = 1 / 3,
Square 1-k ^ 2 = 1 / 9, K ^ 2 = 8 / 9, k > 0,
∴k=2√2/3.

Let o be the origin of coordinates, f be the focus of the parabola y2 = 2x, and a be a point on the parabola. If the projection of the vector FA on the vector of is 1, then the vector o
Let o be the origin of coordinates, f be the focus of the parabola y2 = 2x, and a be a point on the parabola. If the projection of the vector FA on the vector of is 1, what is the vector OA equal to?

Point F (1,0), set point a (x, 2x), then vector FA = (x-1.2x), FA * of = (x-1.2x) (1,0) = X-1, FA * of / | of | = X-1 = 1, then x = 2
Vector OA = (2,4) | OA | = 2 times root sign 5

Given that the parabola y ^ 2 = 4x, M is the moving point on it, O (0,0), f is the focus, then the maximum value of Mo / MF is?

Let m (x, y) MF = x + 1Mo = √ (x ^ 2 + y ^ 2) = √ (x ^ 2 + 4x) Mo / MF = √ (x ^ 2 + 4x) / (x + 1) = √ [(x ^ 2 + 4x) / (x + 1) ^ 2] = √ [(x + 1) ^ 2 + 2 (x + 1) - 3] / (x + 1) ^ 2 = √ [- 3 / (x + 1) ^ 2 + 2 / (x + 1) + 1] = √ - 3 [(1 / (x + 1) - 1 / 3) ^ 2 + 4 / 3], so the maximum value of x = 2 = 2 √ 3 / 3