It is known that the distance from a point a (m, 4) on the parabola C: x ^ 2 = 2PY (P > 0) to its focus is 17 / 4 (1) Finding the value of P and M (2) Let the abscissa of a point P on the parabola C be t (T > 0), the line passing through P intersects C at another point Q, the x-axis intersects m, the line passing through point q is the vertical line of PQ intersecting C at another point n. if Mn is the tangent of C, the minimum value of T is obtained

(1) Point a is on a parabola, so m ^ 2 = 8p,
The Quasilinear equation of parabola is y = - P / 2,
The distance from point a to its focus is equal to the distance to the guide line, so 4 + P / 2 = 17 / 4,
From the above two formulas, we can get: P = 1 / 2, M = 2
(2) The parabolic equation is y = x ^ 2. P point coordinates are p (T, T ^ 2). Suppose Q (x1, X1 ^ 2), m (m, 0), n (X2, X2 ^ 2), then x1, X2, t are different
The results show that the slopes of PM, QM and PQ are equal
t^2/(t-m)=x1^2/(x1-m)=(x1^2-t^2)/(x1-t)=x1+t,
So,
t^2=(t-m)(x1+t) …… ①a
x1^2=(x1-m)(x1+t) …… ①b
The slope of MQ is X1 ^ 2 / (x1-m), the slope of PQ is (x1 ^ 2-T ^ 2) / (x1-t) = X1 + T, and the slope of NQ is (x1 ^ 2-x2 ^ 2) / (x1-x2) = X1 + x2
x1^2/(x1-m)*(x1+x2)=-1,
That is X1 ^ 3 + X1 + X1 ^ 2 * x2 = m ②a
From PQ ⊥ NQ: (x1 + T) * (x1 + x2) = - 1 ②b
For the parabolic equation y = x ^ 2, y '= 2x,
The slope of the tangent of point n on the parabola is 2 * x2,
The slope of line Mn is: x2 ^ 2 / (x2-m),
Because the straight line Mn is tangent to the parabola, so x2 ^ 2 / (x2-m) = 2 * X2, that is,
x2*(2m-x2)=0,…… ③
So x2 = 0 or x2 = 2m
(1) When x 2 = 0, from the formula B, (x 1 + T) * x 1 = - 1,
So t = - x1-1 / x1,
Because t > 0, so X10, so x1

It is known that parabola y & # 178; = 2px (P > 0) and hyperbola X & # 178; (A & # 178; - Y & # 178; (B & # 178; = 1 (a, b > 0) have the same focal point F, point a is an intersection point of two curves, and AF ⊥ X axis. If line L is an asymptote of hyperbola, then the area where the inclination angle of line L lies may be the same
A.(0,π/6) B.（π/3,π/2）
I don't know which one of a and B to choose at the end of the calculation (well, I've worked out both results, I don't know which one to give up)

A: hyperbolic right focus C = √ (A & # 178; + B & # 178;), parabola focus abscissa is p / 2 = √ (A & # 178; + B & # 178;), so p = 2 √ (A & # 178; + B & # 178;). Two curve equations are established to calculate the intersection point. The abscissa of the intersection point is equal to the abscissa of the focus, so: [√ (A & # 178; + B & # 178;)] &# 178 / A & #

What kind of conic is suitable for parametric equation?

You use general equations to do obviously tedious things

It is true to find a, B. LIM (X -- > 0) 1 / (BX SiNx) definite integral [0 -- > x] t ^ 2DT / (a + T) ^ (1 / 2) = 1

Applying lobida's rule, the original formula = LIM (x - & gt; 0) x ^ 2 / √ (a + x) * (b-cosx) because x approaches 0, x ^ 2 approaches 0, and the limit is 1, so b-cosx approaches 0, and B = 1 is substituted into LIM (x - & gt; 0) x ^ 2 / √

Solving a, B. LIM (X -- > 0) 1 / (BX SiNx) definite integral [0 -- > x] t ^ 2DT / (a + T) ^ (1 / 2) = 1 is tenable

Lobida, the numerator x2 / √ a + X, denominator b-cosx, becomes x2 / (b-cosx) √ a, so B = 1, a = 4

Calculate 27 ^ m / 9 ^ m / 3

I'm glad to meet your question
27^m/9^m/3
=3^3m/3^2m/3
=3^(3m-2m-1)
=3^(m-1)

Calculation: 81 ^ m × 27 ^ M-9 & # 178; × 9 ^ m × 3 ^ 5m-4

=3^7m-3^(2+2m+5m-4)
=3^7m-3^(7m-2)
=3^7m-1/9 x 3^7m
=8x3^(7m-2)

Square cardboard with side length of 27, subtract the same square from each corner, and stack the rest into an uncovered cube. What is the volume of this uncovered cube?

27/3=9

As shown in the figure, cut out two rectangles with side length of X along both sides of the square with side length of 4, and the area of the remaining part is 9. The equation can be listed as follows And the solution is X= ．

Let the cut side length be x, then according to the question, it is easy to list the equation as 16 - (4x × 2-x2) = 9, that is, 16-8x + x2 = 9, the solution is x = 1, x = 7 (not in line with the actual situation, rounding off)

In the square with side length x (x > 0), the area of the remaining part is (6x-9) after removing a small square

6x-9 > 0, x > 1.5
The area of the removed small square is x ^ 2 - (6x-9) = x ^ 2-6x + 9 = (x-3) ^ 2,
When x > 3, the side length of small square is x-3; when 1.5