A question about ellipse valuation! Difficult .

The elliptic equation is.

x^2/a^2+y^2/b^2=1

The coordinates of point A can be set as (ACOS, BSIN) and B coordinates as (ACOS, BSIN).

oa vertical ob, so acos * acos + bsin *bsin = 0

Twice with the same addition of cos *cos, get.

a^2+b^2tanα*tanβ=0

tanβ=-(a^2/(b^2*tanα))

1/oa^2+1/ob^2

1/(a^2(cosα)^2+b^2(sinα)^2) +1/(a^2(cosβ)^2+b^2(sinβ)^2)

cosα)^2+(sinα)^2)/(a^2(cosα)^2+b^2(sinα)^2) +cosβ)^2+(sinβ)^2)/(a^2(cosβ)^2+b^2(sinβ)^2)

1+(tanα)^2)/(a^2+b^2(tanα)^2) +1+(tanβ)^2)/(a^2+b^2(tanβ)^2)

1+(tanα)^2)/(a^2+b^2(tanα)^2) +1+((a^2/(b^2tanα))2)/(a^2+b^2((a^2/(b^2tanα))2)

a^2*b^2+a^2*b^2*(tanα)^2+b^4*(tanα)^2+a^4)/(a^2*b^4*(tanα)^2+b^2*a^4)

a^2+b^2)/(a^2*b^2)

So 1 oa 2 + 1 ob 2 is a fixed value (a 2 + b 2) (a 2 * b 2).

Introducing a method of proof of geometry:

Crossing point A is perpendicular to AC to OA to A

Crossing the point b is perpendicular to ob to b

AC crosses BC at point C, connects AB and OC, and AB crosses OC at point D and passes through point A as AH vertical OC at H

Let the area of the rectangle 0acb be s, then.

ob=s/|oa|

oa=s/|ob|

So |oc|^2=|ab|^2

oa|^2+|ob|^2

s 2 (modulus squared of 1 oa + modulus squared of 1 ob) because s 2=|oc|^2|ah|^2

So modulo square of 1 oa + modulo square of 1 ob =|oc|^2/s^21/|ah|^2

So the original proposition is equivalent to proof |ah|is a fixed value.

The following is calculated using the formula for the distance from the point to the line|ah|

a(acos,bsin) b(-asin,bcos) because d is the ab midpoint.

So d(a2(cos-sin), b2(cos+sin)).

Write the equation for the straight line od, which is: b(cos + sin) x + a(cos -sin ) y=0

Substitute a(ACOS, BSIN) into the distance formula to calculate and sort.

ah|=ab root number (a 2 + b 2), which is a fixed value.

So the original proposition holds.

Modulus squared of 1 oa + modulus squared of 1 ob = 1 |ah|^2=(a^2+b^2)/(ab)^2

Convert to sum squared, and then use a special value.

I gave you an answer, pay attention to check.

The first is to find c: we can find it with c 2=a 2-b 2; You draw a diagram on scratch paper and then take point p on the ellipse, connecting pf1 and pf2; (The following is an application of the mean inequality).

by, and pf1+pf2=2a, then the first question is solved.

The second question is made by (pf1+pf2) 2=pf1 +pf2 2+,(pf1+pf2) ,pf1+pf2=2a, when there is a maximum, then there is a minimum on the right, so the result is out hey.

You didn't write the whole topic, but I didn't plan to write the whole step for you, just to say the key, the key is to use the definition of the ellipse |pf1|、|pf2|The relationship between the two is found, that is, |pf1|+|pf2|=2a (the length of the long axis), so these two problems are finally transformed into the range problem of the unary 2nd order polynomial in the interval, and it is easy to get the result, just pay attention to |pf1|AND |pf2|What is the length range.

Upstairs is talking about standard algebraic solutions. The solution to the sophomore math should be to turn the distance to the focal point to the distance to the alignment, and set |pf1|,|pf2|Expressed linearly by x, the quadratic extreme value problem homogenized to x is solved.

Hello Because the calculation is a bit troublesome, the result may not be correct, I didn't check the idea, if you still don't understand the range of parameters, you can contact me, see the figure below.

Ellipse c long axis "2

The detailed process is shown in the figure.

Please, perfect your topic, can you read it through for yourself?

Let ab be any chord of the ellipse x 2 a 2+y 2 b 2=1(a>b>0), m be the midpoint of ab, let the slopes of om and ab exist, and let kom,kabSo what is the relationship between KOM and KAB's group imitation? And prove your conclusions.

Solution: Let x 2 a 2

y 2 b 2 = 1 is the equation for elliptical fiber.

Then: (a 2-c 2) c=1

And: Chord length|ab|=2(a2-c 2) a=root2Upper formula The following formula yields a=croot2

So e = root or foci2 2

1. The ratio of a, b, and c can be obtained by eccentricity, and the elliptic equation with one parameter is set.

2. The relationship between the simultaneous ellipse and the straight line, the relationship between x1, x2, y1, y2 is obtained through Veda.

3. Because the circle with mn as the diameter is over the center of the circle, there is x1x2+y1y2=0 to calculate the parameters in the ellipse.

I only provide ideas, if you want to improve, don't just look at the answers made by others, try it yourself. Hopefully, something will be rewarded.

P.S. The key to this problem is the transformation of the last step, which uses two perpendicular vectors to solve the unknown (the circumferential angle of the radius is a right angle).

a^2-b^2=c^2=20

> (a+b)(a-b)=20

> a-b=2

> a=6 b=4

That is, the standard equation for an ellipse is: x 2 36 + y 2 16 = 1

The abscissa is denoted as x1 and x2 respectively

Vector af = 3 vector fb

f is the fixed score point of the line AB.

x1+3x2)/4=c

That is, x1 + 3x2 = 4c

It is defined by the second of the ellipse.

af|/(-x1+a²/c)=c/a

The focal radius is |af|=-cx1/a+a

The same can be learned|bf|=-cx2/a+a

af|+3|bf|=-c(x1+3x2)/a+4a=-4c²/a+4a=2|af|

k>0 is known

If the inclination angle of the straight line is a, it depends on the image.

a²/c-c+|af|cosa=d1 (1)a²/c-c-|bf|cosa=d2

3(a²/c-c)-3|bf|cosa=3d2 (2) from (1)(2) to obtain 2(a c-c)=2|af|cosa then -4c a+4a=2|af|=2(a c-c) cosa=a 2c=1 3

tan²a+1=1/cos²a

k=tana=√2

It's very easy to figure out the process, and this method is the least computationally intensive.

The analytic expression of the straight line is represented by y, and the value of y1*y2 can be obtained by denoting x in the elliptical formula, and the maximum value can be obtained according to the mean inequality, and it seems that the slope is 0 at its maximum.