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Parabola y = 4x, straight line y = X-1 and parabola intersect at two points a and B, calculate triangle OAB area, O is the origin
Method 1
Y = X-1 through parabola y & # 178; = 4x focus f (1,0)
The line y = X-1 and y = 4x eliminate x simultaneously
We get y & # 178; = 4 (y + 1)
That is Y & # 178; - 4y-4 = 0
Let a (x1, Y1), B (X2, Y2)
Then Y1 + y2 = 4, y1y2 = - 4
The OAB area of triangle
S=SΔAOF+SΔBOF
=1/2*|OF|*(|y1|+|y2|)
=1/2|y1-y2|
=1/2√[(y1+y2)²-4y1y2]
=1/2√[16+16]
=2√2
Method 2
Y = X-1 through parabola y & # 178; = 4x focus f (1,0)
The line y = X-1 and y = 4x eliminate y simultaneously
x^2-6x+1=0
Let a (x1, Y1), B (X2, Y2)
x1+x2=6
|AB|=x1+x2+p=8,
Distance from origin to line = 1 / √ 2
Triangle OAB area
S=1/2*|AB|*1/√2=1/2*8*1/√2=2√2
Given that the line y = x + 2 and the parabola y = x ^ 2 + 2x intersect at two points a and B, and O is the origin of the coordinate, the area of △ AOB is calculated
From the known x + 2 = x ^ 2 + 2x to x ^ 2 + X-2 = 0 to get (x-1) (x + 2) = 0, we can get x = 1 or x = - 2, substitute x = 1 into y = x + 2 to get y = 3, substitute x = - 2 into y = x + 2 to get y = 0, that is, the coordinates of points a and B are (1,3) (- 2,0), that is, the bottom edge of triangle AOB is 2 long and 3 high, so the area of triangle AOB is s = 1 / 2 * 2 * 3 = 3
Given that two points a and B on the parabola y = 2x2 form an isosceles right triangle with the origin o, the coordinates of two points a and B are obtained
Let a (a, 2A2), then | a | = 2A2, ∵ a ≠ 0, | a = ± 12, ∵ a (12, 12), B (- 12, 12)
It is known that the straight line L passes through the focus of the parabola x ^ 2 = 4Y, and intersects with the parabola at two points a and B, and the point O is the origin of the coordinate. It is proved that the angle AOB is an obtuse angle. If the area of the triangle AOB is 4, the equation of the straight line L is obtained
It is known that the straight line L passes through the focus of the parabola X & # 178; = 4Y, and intersects with the parabola at two points a and B, and the point O is the origin of the coordinate. It is proved that the angle AOB is an obtuse angle. If the area of the triangle AOB is 4, the equation of the straight line L is obtained
It is proved that: (1) parabolic parameters: 2p = 4, P = 2, P / 2 = 1, focus f (0,1);
Let the equation of l be y = KX + 1; substituting it into the parabolic equation, we get X & # 178; = 4 (KX + 1), that is, X & # 178; - 4kx-4 = 0; let a (X & # 8321;, Y & # 8321;),
B (X &;, Y &;), then x &; + X &; = 4K, X &; X &; = - 4;
y₁y₂=(kx₁+1)(kx₂+1)=k²x₁x₂+k(x₁+x₂)+1=-4k²+4k²+1=1；
So cos ∠ AOB = OA & ᦇ 8226; ob / [∣ OA ∣ ob ∣] = (X & ᦇ 8321; X & ᦇ 8322; + Y & ᦇ 8321; Y & ᦇ 8322;) / √ [(X & ᦇ 8321; &ᦇ 178; + Y & ᦇ 8321; &ᦇ 178;) (X & ᦇ 8322; &ᦇ 178; + Y & ᦇ 8322; &ᦇ 178;)
=(-4+1)/√[(x₁²+y₁²)(x₂²+y₂²)=-3/√[(x₁²+y₁²)(x₂²+y₂²)
RELATED INFORMATIONS
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- 8. If the line passing through the focus F of the parabola y2 = 4x intersects with the parabola at two points a and B, then OA · ob=______ ．
- 9. The distance from the focus of the parabola y = 4x2 to the collimator is______ ．
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- 13. It's very urgent. The parabola y = ax-3ax + 3 intersects the x-axis at a (- 1,0) B (4,0) intersects the y-axis at C The square of parabola y = ax - 3ax + 3 intersects X axis at a (- 1,0) B (4,0) intersects Y axis at C. 1. Find the analytical formula of parabola 2. Translate the parabola along the Y-axis and intersect the y-axis at point D. when BD = 2CD, find the coordinates of point D 3. If the parabola is folded along the straight line x = m, make the folded parabola intersecting straight line BC at two points P and Q, and P and Q are symmetrical about point C, draw the value graph of m by yourself. Because I don't have a picture
- 14. As shown in the figure, the parabola passes through three points a (4,0), B (1,0), C (0, - 2). (1) find the analytical formula of the parabola (2) There is a point on the parabola above the line AC (3) P is a moving point on the parabola, passing through P as the PM ⊥ X axis, and the perpendicular foot is m. is there a point P, so that the triangle with vertices a, P, and M is similar to △ OAC? If it exists, the coordinates of the qualified point P are requested; if it does not exist, please explain the reason
- 15. In the rectangular coordinate system, there are two intersections between the parabola y = ax ^ 2 + BX + C and the X axis, and the distance from the intersection to the origin is greater than 1 Then the minimum value of ABC is (), and a + B + C = ()
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