ZOJ 3919 Ellipse（數學）

題意:

給定一個標準橢圓方程,x2a2+y2b2=1(a≥b>0)
問有一個圓心在原點(0,0)的內切圓的橢圓內接平行四邊形的,最大和最小面積

分析:

顯然只能內接矩形,設第一象限與橢圓的交點爲(x,y)
則S矩形=2x⋅2y=4xy
x2a2+y2b2=1⇒bx2+ay2=a2∗b2
由基本不等式x2+y2≥2xy得,bx2+ay2≥2abxy
即a2∗b2≥2abxy⇒4xy≤2ab
所以Smax=2ab
最小的是個正方形,我也不知道爲啥,感覺是−−設第一象限與橢圓的交點爲(x,y)
x2a2+x2b2=1⇒(a2+b2)x2=a2∗b2⇒x2=a2b2a2+b2
所以Smin=4x2=4a2b2a2+b2

代碼:

//
//  Created by TaoSama on 2016-02-29
//  Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

double a, b;

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%lf%lf", &a, &b) == 2) {
        double maxs = 2 * a * b;
        double mins = 4 * a * a * b * b / (a * a + b * b);
        printf("%.12f %.12f\n", maxs, mins);
    }
    return 0;
}