ZOJ 3919 Ellipse（数学）

题意:

给定一个标准椭圆方程,x2a2+y2b2=1(a≥b>0)

问有一个圆心在原点(0,0)的内切圆的椭圆内接平行四边形的,最大和最小面积

分析:

显然只能内接矩形,设第一象限与椭圆的交点为(x,y)

则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

最小的是个正方形,我也不知道为啥,感觉是−−设第一象限与椭圆的交点为(x,y)

x2a2+x2b2=1⇒(a2+b2)x2=a2∗b2⇒x2=a2b2a2+b2

所以Smin=4x2=4a2b2a2+b2

代码:

//
//  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;
}