It is known that the standard equation of parabola is Y & # 178; = 8x. Find its focal coordinate and quasilinear equation

It is known that the standard equation of parabola is Y & # 178; = 8x. Find its focal coordinate and quasilinear equation

y²=8x
Focus coordinates (P / 2,0)
Focus coordinates (2,0)
Quasilinear equation: x = - 2
How does arctan ((1 + x) / (1-x)) expand into a power series?
Please be more specific. I'm a little dizzy.
In this paper, we first obtain the derivative, then use Taylor expansion (x) = f (X.) + F '(X.) (X-X.) + F' '(X.) / 2! (X-X.) ^ 2, + F' '' (X.) / 3! (X-X.) ^ 3 + +f(n)(x.)/n!(x-x.)^n+Rn
Let a and B be rational numbers, if the solution of inequality (2a-b) x + 3a-4b0 is
The solution of (2a-b) x + 3a-4b0
-20x -5 >0
X
（2a-b)x+3a-4b0 x
The focal coordinate of parabola X & # 178; = - 2 / 1 y is, and the Quasilinear equation is
How does arctan ((1 + x) / (1-x)) expand into a power series?
This problem has been solved before. 1: if you notice an equation, this problem is relatively simple. Tan (π / 4 + arctanx) = (1 + x) / (1-x), so arctan [(1 + x) / (1-x)] = arctan [Tan (π / 4 + arctanx)] = π / 4 + arctanx, so the original formula = π / 4 + arctanx, so you can directly use the expansion of arctanx to do
It's easy to work out the integral of the result after derivation.... I want detailed steps
Given that AB is a rational number, if the inequality (2a-b) x + 3a-4b "0 and 4-9x are of the same solution, then the inequality
It is known that a and B are rational numbers if the inequality (2a-b) x + 3a-4b
Inequality (2a-b) x + 3a-4b
Solution 4-9x4 / 9
Solution (3a-2b) - (3a-4b) / (3a-4b)
For the convenience of calculation, we say 3a-4b = 4, b-2a = 9 (actually 4K, 9K).
obtain.... Let's do it by ourselves. That's the way
The solution set X4 / 9 of the former inequality;
It's impossible to solve the problem at the same time. Is the topic wrong.
There is a problem with the answer upstairs: X - (3a-4b) / (2a-b)
From (2a-b) x + 3a-4b
The focal coordinate and quasilinear equation of parabola X & # 178; = 4Y are
The focus of parabola X & # 178; = 2PY (P > 0) is f (0, P / 2), and the collimator is y = - P / 2
Then: the focus of parabola X & # 178; = 4Y is f (0,1), and the collimator is y = - 1
There is basically no process
Focal distance P = 2
The focus is on the positive half of the y-axis,
Therefore, the focal point is (0,1) and the Quasilinear equation is y = - 1
Expand f (x) = 1 / (x Λ 2-4x-5) into a power series of X
Write (x) 1 / x = / (x-1 / X-30)
1 / (1 + x) and 1 / (1-x / 5) will expand
Given that a and B are rational numbers, the solution of inequality (2a-b) x + 3a-4b greater than 0 is X & gt; 4 / 9. Find the solution set of inequality (a-4b) x + 2a-3b
If the line y = (2a-b) x + 3a-4b passes through the point (9 / 4,0) and 2a-b & gt; 0, then 6A = 5b, a and B are greater than 0 (a-4b) x + 2a-3b = - (19 / 6) BX - (4 / 3) B, the line y = (a-4b) x + 2a-3b passes through the point (- 8 / 19,0) and the slope is negative (a-4b) x + 2a-3b & gt; 0, the solution is X & lt; - 8 / 19
Finding the focal coordinate and quasilinear equation of parabola y = 2x & # 178
Parabola x ^ 2 = Y / 2 = 2PY, P = 1 / 4
The focus on the y-axis is (0, P / 2), that is, (0, 1 / 8)
The Quasilinear equation is y = - P / 2 = - 1 / 8