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In the plane rectangular coordinate system, O is the coordinate origin, oabc is a rectangle, point a coordinates (10,0) point C coordinates (0,4) point D is the midpoint of OA, point P is on the BC side When the triangle ODP is an isosceles triangle with waist length 5, the coordinate of point P is?
The coordinates of point P are (8,4)
If the left focus F of the ellipse x ^ 2 + 2Y ^ 2 = 4 makes a chord AB with an inclination angle of 60 degrees, then the length of the chord AB is
Let the equation passing through the right focus be y = radical 3 (x-radical 2)
By combining the elliptic equations, we find that | x1-x2 | = 8 / 7,
So the length of AB is 16 / 7
It is known that: as shown in the figure, O is the coordinate origin, the quadrilateral oabc is a rectangle, a (10,0), C (0,4), point D is the midpoint of OA, and point P moves from C to B at the speed of 1 unit per second on BC. (1) find the functional relationship between the area s of trapezoidal odpc and time T. (2) when t is the value, the quadrilateral podb is a parallelogram? (3) Whether there is a point Q on the line segment Pb, so that odqp is a diamond. If there is, find the value of T, if not, explain the reason. (4) when △ OPD is an isosceles triangle, find the coordinates of point P
(1) According to the area formula of trapezoid, we get that s = (T + 5) × 42 = 2T + 10 (2) ∵ quadrilateral podb is parallelogram, ∵ Pb = od = 5, ∵ PC = 5, ∵ t = 5 (3) ∵ odqp is diamond, ∵ od = OP = PQ = 5, ∵ in RT △ OPC, we get PC = 3 ∵ t = 3 (4) by Pythagorean theorem when p1o = od = 5
Through the right focus F2 of the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1, make a straight line with an inclination angle of pi / 4 to intersect the ellipse with two points a and B. find: 1) the length of the chord ab
2) The area of triangle AOB (3) is the distance from the left focus F1 to the midpoint of chord ab
The solution: 1) the meaning of the problem, C = √ ^ 2-B ^ 2 = 1, k = Tan π / 4 = 1 of 〈 F2 (1,0)
The linear equation is y-0 = 1 (x-1), which is y = X-1
For the generation of y = X-1 to ellipse x 2 / 4 + y 2 / 3 = 1 with simplified finishing 7 times ^ 2-8 times speed 8 = 0
| AB | =√1 +1 ^ 2 *√（8/7）^ 2-4 * -8 / 7 = 24/7
(2) The linear distance of easy question F1 d = | - 1-1 | / √ 1 ^ 2 + (- 1) ^ 2 = √ 2
∴S△AF1B = 1 / 2 * 24/7 *√2 = 12√2/7
③ 7x ^ 2-8x-8 = 0 has X1 + x2 = 8 / 7 abscissa chord AB midpoint (1 degree + × 2) / 2 = 4 / 7
X1-1 of Y1 = and - 1 of y2 = X2, Y1 + y2 = × 1 + × 2-2 = 10 - 6 / 7
So the vertical coordinates of the midpoint of the chord AB (Y1 + Y2) / 2 = - 3 / 7
The coordinates of the midpoint of AB are (4 / 7, - 3 / 7)
In the right angle trapezoid oabc, CB ‖ OA, ∠ COA = 90 °, CB = 3, OA = 6, Ba = 3, 5
In right angle trapezoid oabc, CB ‖ OA, ∠ COA = 90 °, CB = 3, OA = 6, Ba = 3, 5
The plane rectangular coordinate system as shown in the figure is established by taking the line of OA and OC as the x-axis and y-axis respectively
(1) Find the coordinates of point B;
(2) Given that D and E are the points on OC and ob, OD = 5, OE = 2EB, and the line de intersects the x-axis at point F, the analytical formula of the line De is obtained;
.
In the case of RT △ ABP, ∠ APB = 90 ° AP = 3, ab = 3 root sign 5 BP = (3 root sign 5) &# 178; - (3) &# 178; = 6 ℅ B (3,6) as eg ⊥ X axis at point G, then eg ∥ BH, ∥ OEG ∥ obh, (4 points) ∥ OE / ob = og /
Let F 1 and F 2 be the two focal points of the ellipse X & # 178 / 9 + Y & # 178 / 25 = 1. If AB is a chord passing through the center of the ellipse, find the maximum area of △ F 1 ab
If F2 is connected with a and B respectively, it is easy to know that f1af2b is a parallelogram
S△F1AB=1/2*SF1AF2B=S△F1F2A
In △ f1f2a, the bottom edge F1F2 = 2 * √ (25-9) = 8
When point a is the intersection of ellipse and x-axis, the height h of △ f1f2a takes the maximum value of 3
At this time, s △ f1f2a = 1 / 2 * 8 * 3 = 12
So the maximum value of s △ f1ab is 12
In the rectangular trapezoid oabc, CB ‖ OA, COA = 90 °, OE = 2EB, CB = 3, OA = 6, Ba = 35, OD = 5. The plane rectangular coordinate system as shown in the figure is established by taking the line of OA and OC as x-axis and y-axis respectively
It is proved that: through point B, BG ⊥ X axis intersects X axis at point G, ∵ CB ∥ OA, ∠ COA = 90 °, CB = 3, ≁ og = 3, ≁ GA = oa-og = 6-3 = 3, BG ⊥ X axis, ≁ in right triangle AGB, BG2 = ab2-ga2 = (35) 2-32 = 36, ≁ BG = 6, then ob = 35 is obtained according to Pythagorean theorem, OE = 2 is obtained from OE = 2be
The left and right focus of ellipse x * 2 / 25 + y * 2 / 16 are F1, F2, and the chord AB passes F1
If the circumference of the inscribed circle of the triangle abf2 is pie, and the coordinates of a and B are (x1, Y1) and (X2, Y2), then the absolute value of y2-y1 is? A6 / 3 B10 / 3 C20 / 3 D5 / 3
Who can help to analyze, a good answer, there are bonus points, thank you
A prior theorem is that the area of a triangle is equal to the product of the half perimeter and the radius of the inscribed circle
The radius of inscribed circle is r = π / 2, π = 1 / 2
Triangle perimeter L = 2 * 2A = 20
So s = 1 / 2 * 20 * 1 / 2 = 5
And S = 1 / 2 * | F1F2 | * | y2-y1 | = 1 / 2 * 6 * | y2-y1 | = 5
|y2-y1|=5/3
Choose D
D
In the plane rectangular coordinate system, the quadrilateral oabc is a reminder, OA is parallel to CB, the coordinates of point a are (6,0), and the coordinates of point B are (3,4)
In the plane rectangular coordinate system, the quadrilateral oabc is a trapezoid OA parallel CB, the coordinate of point a is 〈 6,0 〉, the coordinate of point B is {3,4], point C is on the y-axis, the moving point m moves on the OA side, starting from point O to point a; the moving point n moves on the AB side, starting from point a to point B, two moving points, starting at the same time, the speed is 1 unit length per second, when one of them reaches the end, the other stops immediately, Let the motion time of two points be T seconds
1: When t is, Mn is parallel to OC?
2: When t is sum, Mn ⊥ AB?
(1) Coordinates of trapezoid vertex: O (0,0), a (6,0), B (3,4), C (0,4)
OA = 6
AB = √[(3-6)² + (4-0)²] = 5
Let m (T, 0), t ∈ [0,6] be parallel to y-axis, then om = an = t
M. The abscissa of n is the same, let n (T, n)
The equation of AB is: (Y - 0) / (x - 6) = (4 - 0) / (3 - 6) = - 4 / 3
y = -4(x-6)/3
n = -4(t - 6)/3
OM² = AN²
t² = (t - 6)² + (n - 0)² = (t - 6)² + 16(t - 6)²/9 = 25(6-t)²/9
t = ±5(6-t)/3
T = 15 (> 6, rounding off)
t = 15/4
(2)
The slope of AB is k = (4 - 0) / (3 - 6) = - 4 / 3
When Mn ⊥ AB, the slope of Mn is - 1 / k = 3 / 4
Let m (T, 0), t ∈ [0,6], then n (P, 4 (6-P) / 3)
OM² = AN²
t² = (p - 6)² + 16(6 - p)²/9 = 25(6 - p)²/9 (a)
[4(6-p)/3 - 0]/(p - t) = 3/4 (b)
From (a) (b): T = 9 / 4 (another value t = 9 > 6, rounding off)
1、t=3.75
2、t=2.25
three hundred and forty-five thousand three hundred and forty-five
If a line passing through the left focus F of the ellipse X29 + Y25 = 1 and not perpendicular to the x-axis intersects the ellipse at two points a and B, and the vertical bisector of AB intersects the x-axis at point n, then | NF | ab|=______ .
Let a (x1, Y1) and B (X2, Y2), then X1 + x2 = 187, Y1 + y2 = - 107, x1x2 = - 914, the midpoint coordinates of AB are (97, - 57), then the middle perpendicular equation of AB is y + 57 = - (x-97), let y = 0, then x = 47, the coordinates of n (47, 0).. | NF | = (47 − 2) 2 = 107, | ab | = 2 [(187) 2 − 4 × (− 914)] = 307, | NF | ab | = 13
RELATED INFORMATIONS
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