(1) equation for finding line l2
Answer:-1x/3-(22/9)
Solution: y'=2x+ 1
The tangent slope at the point (1, 0) =3, while: l 1⊥l2,
So the slope of l2 =- 1/3.
2x+ 1=- 1/3
x=-2/3
Corresponding y = (-2/3) 2+(-2/3)-2 =-20/9.
L2 crossing point (-2 March,-20 September)
So: the equation of l2: y+(20/9)=(- 1/3)(x+(2/3))
y=-( 1/3)x-(22/9)
(2) Find the area of the triangle surrounded by the straight line l 1, l2 and X axis.
Answer: 125/ 12
Solution: equation l 1: y=3x-3.
Where y=-( 1/3)x-(22/9),
De: x= 1/6，y=-5/2。
And: the intersection of l2 and X axis (-22/3,0)
Area = (1/2) (1-(-22/3)) * (5/2) =125/12.
2, let b≥0, parabola c1:y = x 2+bx-a 2 and parabola C2: y =-x 2+ax+(b √ 2/4) be perpendicular to each other at their intersections.
(1) Find the parabolic equation f(a, b)=0 for point P(a, b).
Answer: f (a, b) = 2a 2+(b √ 2/2)-ab-1.
Solution: parabola C 1:
y'=2x+b
Parabola C2:
y'=-2x+a
(2x+b)(-2x+a)=- 1
4x^2-2(a-b)x-ab- 1=0-( 1)
And: x 2+bx-a 2 =-x 2+ax+(b √ 2/4)
4x^2-2(a-b)x-(2a^2+(b√2/2))=0-(2)
( 1)-(2):
2a^2+ (b√2/2)-ab- 1=0
f(a,b)=2a^2+ (b√2/2)-ab- 1
(2) If the point Q(m, n) moves in the closed area (including the boundary) surrounded by curves x=0, y=0 and f(x, y)=0, find the value range of 3m+n..
Answer: [0,7 √ 2/2]
Solution: f (x, y) = 2x 2+(y √ 2/2)-xy- 1 = 0.
√2(2x^2- 1)-y((√2)x- 1)=0
((√2)x- 1)(2x+√2-y)=0
Therefore, f(x, y)=0 becomes two straight lines:
(√2)x- 1=0, and this line intersects with y=0 at (√ 2/2,0).
2x+√2-y=0, and this line intersects with x=0 at (0, √2).
The intersection of these two straight lines is (√2/2, 2√2).
Now we need a range of 3 m+n.
Can be a straight line with a slope of -3,
As can be seen from the figure:
At point (√2/2, 2√2), 3m+n is the largest, and 3m+n is the largest = (3/2) √ 2+2 = 7 √ 2/2.
At point (0,0), 3m+n is the smallest, and 3m+n is the smallest =0.
So the value range of 3m+n is [0,7 √ 2/2].
3. A hemisphere with a radius of 5 is inscribed with a cuboid whose bottom surface is twice its width (the bottom surface coincides with the bottom surface of the hemisphere), and the maximum volume of the cuboid is found.
Answer: 400√3/27
Solution: let a cuboid be x, 2x, h.
Then: h 2+x 2+(x/2) 2 = r 2
H^2=R^2-(5/4)x^2
(Volume) 2 = V2 = (x * 2x * h) 2 = 4x4 * h 2 = 4x4 (R2-(5/4) x 2)
=4*(64/25)*(5/8)x^2*(5/8)x^2*(r^2-(5/4)x^2)
And: (5/8) x 2+(5/8) x 2+(r 2-(5/4) x 2 = r 2 = fixed value.
When: (5/8) x 2 = (r 2-(5/4) x 2) = r 2/3.
Maximum value of V 2 = 4 * (64/25) * r 6/27
V max = 2 * (8/5) r 3/√ 27 = 400 (√ 3)/9
I don't know why it's different from your answer.
At this time: (5/8) x 2 = r 2/3.
x=√(40/3)，2x=2*√(40/3)，
H^2=R^2/3，H=5/√3
V=x*(2x)*H=2*(40/3)*5/√3
=400(√3)/9
There's no problem coming back like this,
Please look at your answer again.
4. curve y =1-x 2 (x >; 0) is the tangent of the curve, which intersects with the X axis and the Y axis at points M and N respectively. Try to determine the coordinates of p to minimize the area of △MON.
Answer: (√3/3, 2/3)
Solution: y'=-2x
Let the coordinates of point P be (t, 1-t 2).
Then: the equation of MN: y-(1-t2) =-2t * (x-t) =-2tx+t2+1.
M-point coordinates ((t 2+ 1)/(2t), 0), and n-point coordinates (0, t 2+ 1).
△ mon = s = (1/2) ((T2+1)/(2t)) * (T2+1)
=( 1/4)*(t^2+ 1)^2*( 1/t)
When S'=0, s has a maximum value (which can be proved as a minimum value).
s'=( 1/4)*2(t^2+ 1)*(2t)/t -( 1/4)(t^2+ 1)^2/t^2
=( 1/4)(t^2+ 1)(3t^2- 1)/t^2=0
3t^2- 1=0
T=√3/3 (the condition of the topic is t>0)
At this point, 1-t 2 = 2/3.
Coordinates of point P (√3/3, 2/3)