If a and B are two points on the parabola y & # 178; = 3x, a (2, √ 6), O is the vertex, OA ⊥ ob, find the length of ab

Let B (y ^ 2 / 3, y) y

When k is any real number, the curve where the vertex of the parabola y = ½ (x-k) ² + K & #178; is ()
A,y=x² b,y=-x² c,y=x² (x>0) d,y=-x² (x>0)

From the equation of parabola y = ½ (x-k) ² + K & #178; it can be seen that the parabola is characterized by opening upward, the axis of symmetry is x = k, and the minimum value is K & #178;
From this we can see that when k is any real number, K & # 178; ≥ 0, the answer is obvious, a, y = x & # 178;

It is known that the parabola y2 = - X and the straight line L: y = K (x + 1). If the parabola and the straight line L intersect at two points AB, O is the vertex of the parabola

Solution
Y ^ 2 = - X and y = K (x + 1) are combined to get k ^ 2x ^ 2 + (2k ^ 2 + 1) + K ^ 2 = 0. According to Weida's theorem, we get x1x2 = 1. Y1 ^ 2Y2 ^ 2 = x1x2. Because y1y2 is a different sign, y1y2 = - 1, so x1x2 + y1y2 = 1-1 = 0. The coordinates of vector OA are (x1, Y1) and ob are (X2, Y2), so oaob is vertical

The straight line y = X-2 and the parabola y2 = 2x intersect at two points a and B, and prove that OA ⊥ ob (o is the coordinate origin)

It is proved that y2-2y-4 = 0 ∧ Y1 + y2 = 2, y1y2 = - 4 ∧ x1x2 = (Y1 + 2) (Y2 + 2) = y1y2 + 2 (Y1 + Y2) + 4 = 4 ∧ y1y2x1x2 = - 1, that is, (Y1 / x1) (Y2 / x2) = - 1koa = y1x1, kob = y2x2 ∧ KOA · kob = y1y2x1x2 = - 1 ∧ OA ⊥ ob

12. Given that the line y = X-2 intersects the parabola y * 2 = 2x at points a and B, (1) prove OA ⊥ ob (2) find the length of ab
12. Given that the line y = X-2 intersects the parabola y * 2 = 2x at points a and B, (1) prove OA ⊥ ob (2) find the length of ab
emergency

If y = X-2 and the parabola y * 2 = 2x intersect at points a and B, then Xa and XB are the two roots of (X-2) ^ 2 = 2x
x^2-4x+4=2x
X = 3 ± root 5
Xa = 3-radical 5, Ya = 1-radical 5
XB = 3 + radical 5, Yb = 1 + radical 5
k1=yA / xA,k2=yB / xB
K1 * K2 = yayb / xaxb = (1-radical 5) (1 + radical 5) / (3 + radical 5) (3-radical 5) = - 1, so OA ⊥ ob
|Ab | = radical [(XB XA) ^ 2 + (Yb ya) ^ 2] = radical [(2 radical 5) ^ 2 + (2 radical 5) ^ 2] = 2 radical 10

The straight line y = X-2 and the parabola y2 = 2x intersect at two points a and B, and prove that OA ⊥ ob (o is the coordinate origin)

It is known that a straight line passing through point P (2,0) with a slope of 4 / 3 and a parabola y ^ 2 = 2x intersect at two points a and B. let the midpoint of line AB be m. find the mark of point M
If I do it with a linear parametric equation,
x=2+3/5t
y=4/5t
After substitution, 8t ^ 2-15t-50 = 0
How to find the M coordinate at this time

Let a (x1, Y1), B (X2, Y2), then M ((x1 + x2) / 2, (Y1 + Y2) / 2)
Using the parametric equation, m (2 + 3 / 5 * (T1 + T2) / 2,4 / 5 * (T1 + T2) / 2)
From 8t ^ 2-15t-50 = 0 and WIDA's theorem, we get that T1 + T2 = 15 / 8,
So m (41 / 16,3 / 4)

It is known that the straight line passing through point P (2,0) with a slope of 4 / 3 and the parabola y = 2x intersect at two points a and B. let the midpoint of the line segment be m. calculate the coordinates of point M

From the straight line passing through point P (2,0), the slope is 4 / 3, the linear equation is
x=t+2
y=4/3t
Substituting into the parabola y ^ 2 = 2x, we get
16/9t^2=2*(t+2)
Simplify
8t^2-9t-18=0
So as to get
t1+t2=9/8
The coordinates of the midpoint m of line AB are
(x1+x2)/2=(t1+t2)/2+2=9/16+2=41/16
(y1+y2)/2=4/3*(t1+t2)/2=3/4.

It is known that: a straight line and a parabola passing through point P (2,0) with a slope of 4 / 3: the square of y = 2x intersect at two points a and B. let the midpoint of line AB be m, and calculate the coordinate of point M

y = 4(x - 2)/3
y*y = 2x
2x = 16(x - 2)^2/9
8x^2 - 32x + 32 - 9x = 0
8x^2 - 41x + 32 = 0
x1 + x2 = 41/8
Then M coordinates (x, y), x = (x1 + x2) / 2 = 41 / 16
Y = 4(X-2)/3 = 3/4

The value of K is ()
A. - 1 or 2B. 2C. - 1D. 1 + 3

Let a (x1, Y1), B (X2, Y2). From y = KX − 2Y2 = 8x, we can get k2x2 - (4K + 8) x + 4 = 0, from △ = [(4K + 8)] 2-16k2 = 64K + 64 > 0, we can get k ＞ - 1; X 1 + x 2 = 4K + 8K 2. The abscissa of a and B midpoint is 2, ∩ 4K + 8K 2 = 4, and the solution is k = - 1 (rounding) or K = 2. Therefore, if the line y = kx-2 and parabola y 2 = 8x intersect at a and B, and the abscissa of AB midpoint is 2, the value of K is 2

If a and B are two points on the parabola y & # 178; = 3x, a (2, √ 6), O is the vertex, OA ⊥ ob, find the length of ab