Two tangent lines of the parabola y = (X-H) &# + K (k > 0) are introduced through the origin o of the coordinate. The tangent points are a and B1 respectively. It is proved that the line segment AB is bisected by the y-axis 2 find the slope of line ab

Two tangent lines of the parabola y = (X-H) &# + K (k > 0) are introduced through the origin o of the coordinate. The tangent points are a and B1 respectively. It is proved that the line segment AB is bisected by the y-axis 2 find the slope of line ab

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(1) proof: let a (x1, Y1), B (X2, Y2)
∵y=(x-h)^2+k(k>0)
∴y’=2(x-h)
The slope of the tangent passing through point a is equal to the slope of Ao
The tangent slope passing through point B is equal to the Bo slope
∴2(x1-h)=y1/x1
2(x2-h)=y2/x2
That is, 2x1 (x1-h) = (x1-h) ^ 2 + K ①
2x2(x2-h)=(x2-h)^2+k…………………… ②
From ① to ②
2[(x1+x2)(x1-x2)-h(x1-x2)]=(x1+x2-2h) (x1-x2)
Divide both sides by (x1-x2)
2[(x1+x2)–h]= (x1+x2-2h)
That is, X1 + x2 = 0
That is, the line AB is bisected by the Y axis
⑵kAB=(y1-y2)/ (x1-x2)
From (1), we get that KAB = (1 - 2) / (x1-x2) = 2 [(x1 + x2) - H] and X1 + x2 = 0
∴kAB= -2h

As shown in the figure, after the parabola y = − 12x2 is translated, it passes through the origin O and point a (6, 0). The vertex of the translated parabola is point B, and the axis of symmetry intersects with the parabola y = − 12x2 at point C. then the area of the shadow part enclosed by the line BC and two parabolas in the figure is ()
A. 212B. 12C. 272D. 15

When x = 3, y = - 12 × 32 = - 92, and the coordinates of point C are (3, - 92), passing through point C as CD ⊥ Y axis at point D. according to the symmetry of the parabola, the area of the shadow is equal to the area of the rectangular CDOE, s = 3 ×| - 92 | = 272

Translate the parabola y = XX + 2x-8, make it pass through the origin, and write an analytical expression of the parabola after translation

y=(x+1)^2