The equation of conic curve. Find the hyperbolic equation with two vertices of ellipse x squared / 16 + y squared / 9 = 1 as focus and ellipse focus as vertex

Elliptic equation: x ^ 2 / 16 + y ^ 2 / 9 = 1, that is, a = 4, B = 3 = = > 4 ^ 2-3 ^ 2 = 7 (a ^ 2-B ^ 2 = C ^ 2),
Two focal points (- √ 7,0), (√ 7,0) are obtained
The two vertices of the ellipse are the focus and the focus is the vertex
So the hyperbolic equation a = √ 7, B = 3
Hyperbolic equation: x ^ 2 / 7-y ^ 2 / 9 = 1

What's the difference between ellipse, hyperbola and parabola

The uniform property of conic guide line: the ratio of the distance between any point on conic curve and the distance from its corresponding guide line is eccentricity E

When a is a value, the curve y = ax ^ 2 is tangent to y = LNX? I am stupid,

If the tangent point is p (x, y), the following conclusions can be drawn
(1) The point P is on both y = ax ^ 2 and y = LNX, so y = ax ^ 2 = LNX
(2) The slopes of the tangent lines of the two curves at point P are equal, that is, the derivatives of the functions y = ax ^ 2 and y = LNX at X are equal. Therefore, 2aX = 1 / X
The simultaneous equations are: y = ax ^ 2 = LNX, 2aX = 1 / x, and a = 1 / (2e) is obtained

Conic and derivative SOS assault It's time to take a math exam at the beginning of school. Take 1-1, a whole book of conic curve and derivative. What about the two mountains? The head teacher told me before the holiday that we should take a serious tutorial, but it's easy to say and it's difficult to do it. The cram class has no course about the scope of the examination, so we have to review it at home. No one asked = = =, how can we attack these two mountains on the afternoon of the 20th Try not to copy the answers in class, listen carefully, go home and do questions carefully. I have only three days left. I want practical methods [or useful materials] Here to blame yourself: young people do not work hard, old man sad, you must not learn from me!

There are still three days. First use one day to memorize the definition and formula. The formula of conic curve can be linked together. It is very regular. As for the derivative, remember the method to get the derivative. The remaining two days will be a surprise to the topic. Do not delay the time of each problem too long. You will not refer to the answer immediately. Try to do the basic problem as much as possible

1. Monotone increasing interval of function f (x) = 2x ^ 2-inx + 1 2. The number of roots of equation 2x ^ 3-6x ^ 2 + 7 = 0 in interval (0,2) 3. The minimum value of function y = in ^ 2x / X 4. If the maximum value of the function y = - x ^ 2-2x + 3 on the interval [a, 2] is 15 / 3, then a = 5. Let x ≥ 0, y ≥ 0. X + 3Y = 9. Find the maximum value of x ^ 2 * y (x squared by Y, y is not an index) Wrong number 4 If the maximum value of the function y = - x ^ 2-2x + 3 on the closed interval of interval (a, 2) is 15 / 4, then a = Do you have a more detailed process for 3 and 5

2. Let f (x) = 2x ~ 3-6x ~ 2 + 7, its function image is like a capital letter N, in mathematics textbooks, the geometric meaning of derivative is discussed

Given that the function f (x) = ax ^ 3 + BX ^ 2-3x obtains the extremum at x = ± 1, we try to discuss whether f (1) and f (- 1) are the maximum or minimum of the function f (x)?

f(x)=ax^3+bx^2-3x
f'(x)=3ax^2+2bx-3
If the function obtains the extreme value at x = ± 1, the derivative at this point is 0;
F '(1) = 3A + 2b-3 = 0, f' (- 1) = 3a-2b-3 = 0, we can get: a = 1, B = 0;
So: F (x) = x ^ 3-3x
In this case, f '(x) = 3x ^ 2-3,
-10, which is an increasing function;
Therefore, at x = - 1, it is the maximum, f (- 1) = 2;
X = 1 is the minimum, f (1) = - 2

Let f (x) be a cubic function whose image is symmetric about the origin. When x = 1 / 2, the minimum value of F (x) is -1. Find the analytic expression of function f (x)

When f (x) = ax? + MX? + BX + n is symmetric about the origin, the odd function f (- x) = f (x), so m = n = 0f (x) = ax? + bxf '(x) = 3ax? + BX = 1 / 2, the function value is - 1F (1 / 2) = - 1A / 8 + B / 2 = - 1 simultaneous 3A / 4 + B = 0A = 4, B = - 3, so f (x) = 4x & S

High and middle derivatives and tangent problems In {an}, A1 = 1, A2012 = 4, function f (x) = x (x-a1) (x-a2) (x-a2012), then the tangent equation of curve y = f (x) at point (0,0) is_______ Thank you for your advice~

In addition, G (x) = (x-a1) (x-a2) In the (x-a2012) = = > F (x) = XG (x) = > F '(x) = g (x) + XG' (x) = > F '(0) = g (0) = A1A2... A2012 in the equal ratio sequence {an}, A1 = 1, A2012 = 4 = = > A2012 = q ^ 2011 = 4f' (0) = QQ ^ 2... Q ^ 2011 = q ^ (2011 * 1006) = 4 ^ 1006y = f (x) tangent at point (0,0)

On tangent and derivative I have a question, for example, if a quadratic function f (x) = x ^ 2, then its derivative at point (x, f (x)) is f '(x) = 2x, that is to say, the slope of tangent line of any quadratic function passing through the point (x, f (x)) is 2x. Why? The derivative function f' (x) = 2x is a limit, which is infinitely close to 2x, but it can never reach 2x. But why say that any point (x, f (x)) passes through a point (x, f (x)) has a limit, The slope of tangent line of quadratic function of F (x)) is 2x. I have proved that the slope of tangent line of quadratic function passing through point (x, f (x)) is 2x, which is true Ask the master to explain

Well Even the concept of limit is not clear
The limit is a definite value! In the sequence {an} = 1,1 / 2,1 / 3,1 / 4 The limit is 0! Not your imagination infinitesimal, do not understand, please go to Baidu Encyclopedia reference definition, do not know you have calculus?
You say that "derivative function f '(x) = 2x is a limit, it is only infinitely close to 2x, but it can never reach 2x. But why say that the slope of tangent line of any quadratic function passing through point (x, f (x)) is 2x". That is to say, we don't understand the meaning of limit (derivative is a kind of limit)
"I have also proved that there is only one intersection point between a straight line and a hyperbola". This sentence is somewhat ignorant. Tangent line does not necessarily have only one intersection point with a straight line. Y = x ^ 3. Are you a high school student or a college student?

Derivative tangent problem If the straight line passing through the point (0,2) is tangent to the curve y = x ^ 3 and the curve y = x ^ 2 + MX + 2, find M

Y = x ^ 3, y '= 3x? The tangent line passing through (a, a? 3) is y = 3A? (x-a) + a
The tangent passes through (0,2), i.e. 2 = 3A (0-A) + a? = - 2A?, a = - 1
Tangent: y = 3 (x + 1) - 1 = 3x + 2, slope 3
By substituting y = x ^ 2 + MX + 2, y '= 2x + m, y' = 3, x = (3-m) / 2, y = 17 / 4 - M 2 / 4
In the tangent equation:
17/4 - m²/4 = 3((3-m)/2) + 2
M = 3

The equation of conic curve. Find the hyperbolic equation with two vertices of ellipse x squared / 16 + y squared / 9 = 1 as focus and ellipse focus as vertex