(2) according to the tangent length theorem, EA=ED, ∠AEM=∠DEM ∴△EAM≌△EDM let the area of quadrilateral EAMD be s.
The area of the triangle EAM is S 1, so S=2S 1=4√3 ∴S 1=2 √3.
Let E(- 1, b), then s1=1/2 * am * ea = b ∴ e (-1,2 √ 3).
E (- 1, 2 √ 3) and M (1, 0) give Me: Y =-√ 3x+√ 3.
Let the straight line ad: y = kx+b, then k= √3/3, and substitute it into A(- 1, 0) to get the straight line ad: y = √ 3/3 * (x+ 1), and the circle equation: (x- 1) 2.
2(3) Let E(- 1, b), N(x, y), the area of quadrilateral EAMD is S, the area of triangle ADN is S 1, and the area of triangle AMN is S2, then S = 2 *1/2 * Ma * AE = 2b =
,∴d( 1+√(4-b^2),b),n( 1-√(4-b^2),-b)
∴ED is parallel to the x axis and tangent to the circle. ,∴b=2，E(- 1，2)，D( 1，2)，N( 1，-2)，
Let P(x, 2) be substituted into the parabolic equation: y = x 2-2x-3 to get p (1+√ 6,2) or (1-√ 6,2).
2.( 1) Substitute D (2 2,0) and C( 1, -3) into y = ax 2+c, and get: y = x 2-4.
(2) Connect the Y axis of BD to M 1, and get BD from B(- 1,-3) d (2,0): y = x-2, ∴ m1(0,2).
∵m 1A+m 1B = m 1D+m 1B = BD
Choose another point m on the y axis, then ma+MB = MD+MB > BD，
∴M 1(0,-2) is what you want.
(3) Let P(x, y),
S△PAD= 1/2*AD*∣y∣=2*∣y∣，s△abm =( 1+2)* 3/2- 1/2 * 2 * 2- 1/2 * 1 * 1 = 2
∫s△pad = 4s△abm∴∣y∣= 4∫ The extreme point is (0,-3) ∴ y = 4x = 2 ∣ 2.
∴P(2√2,4) or (-2 √ 2,4)
3( 1)∫a( 1，0)，B(0，3) ∴AB:y=-3x+3
(2)∵AB=3∴C(-2,3). Substitute A, B and C into Y = AX 2+BX+C respectively to get Y =-X 2-2x+3 and vertex E(- 1, 4).
(3) straight line ef: y = 4, g (3,0), GH: y =-3x+9 ∴ h (5/3,4)
Let P(x, y), then s △ PAG =1/2 * ag * ∣ y ∣ = ∣ y ∣; s△PEH = 1/2 * EH *(4-y)= 4/3 *(4-y)
∫S△PAG = 3/4S△PEH =- 1√2
Substitute ∴y=2 into the parabolic agenda equation and get x =- 1 √ 2.
∴P is (-1+√ 2,2) or (-1-√ 2,2).