2023浙江大學(xué)強基數(shù)學(xué)逐題解析（4）

2023-06-20 21:44 作者:CHN_ZCY 0人讀過 | 我要投稿

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作畫：WERI

https://www.pixiv.net/artworks/96196782

9. 已知橢圓 $%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1%5Cleft(a%3Eb%3E0%5Cright)$ ，過右焦點 $F$ 作相互垂直的弦 $AC$ ， $BD$ ，已知四邊形 $ABCD$ 的面積的取值范圍為 $%5Cleft%5B8%2C%5Cfrac%7B25%7D%7B2%7D%5Cright%5D$ ，則 $%5Cfrac%7Ba%7D%7Bb%7D%3D$ ___________.

答案? 2

解析? 設(shè) $FA$ 與 $x$ 軸正方向的夾角（逆時針為正）為 $%5Ctheta%20%5Cin%20%5Cleft%5B0%2C2%5Cpi%5Cright)$ ，記 $c%3D%5Csqrt%7Ba%5E2-b%5E2%7D$ 則

$%5Cleft%7CFA%5Cright%7C%3D%5Cfrac%7Bb%5E2%7D%7Ba%2Bc%5Ccos%5Ctheta%7D$

$%5Cleft%7CFC%5Cright%7C%3D%5Cfrac%7Bb%5E2%7D%7Ba-c%5Ccos%5Ctheta%7D$

所以

$%5Cleft%7CAC%5Cright%7C%3D%5Cfrac%7B2ab%5E2%7D%7Ba%5E2-c%5E2%5Ccos%5E2%5Ctheta%7D$

同理

$%5Cleft%7CBD%5Cright%7C%3D%5Cfrac%7B2ab%5E2%7D%7Ba%5E2-c%5E2%5Csin%5E2%5Ctheta%7D$

得

$%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%7D%2B%5Cfrac%7B1%7D%7B%5Cleft%7CBD%5Cright%7C%7D%3D%5Cfrac%7B2a%5E2-c%5E2%7D%7B2ab%5E2%7D%3D%5Cfrac%7Ba%5E2%2Bb%5E2%7D%7B2ab%5E2%7D$

于是

$%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%5Ccdot%5Cleft%7CBD%5Cright%7C%7D%3D%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%7D%5Cleft(%5Cfrac%7Ba%5E2%2Bb%5E2%7D%7B2ab%5E2%7D-%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%7D%5Cright)$

其中

$%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%7D%5Cin%5Cleft%5B%5Cfrac%7B1%7D%7B2a%7D%2C%5Cfrac%7Ba%7D%7B2b%5E2%7D%5Cright%5D$

所以

$%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%5Ccdot%5Cleft%7CBD%5Cright%7C%7D%5Cin%5Cleft%5B%5Cfrac%7B1%7D%7B4b%5E2%7D%2C%5Cfrac%7B%5Cleft(a%5E2%2Bb%5E2%5Cright)%5E2%7D%7B16a%5E2b%5E4%7D%5Cright%5D$

因此

$S_%7B%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%7CAC%5Cright%7C%5Ccdot%5Cleft%7CBD%5Cright%7C%5Cin%5Cleft%5B%5Cfrac%7B8a%5E2b%5E4%7D%7B%5Cleft(a%5E2%2Bb%5E2%5Cright)%5E2%7D%2C2b%5E2%5Cright%5D$

得

$%5Cbegin%7Bcases%7D%0Aa%3Eb%3E0%5C%5C%0A%5Cfrac%7B8a%5E2b%5E4%7D%7B%5Cleft(a%5E2%2Bb%5E2%5Cright)%5E2%7D%3D8%5C%5C%0A2b%5E2%3D%5Cfrac%7B25%7D%7B2%7D%0A%5Cend%7Bcases%7D$

即 $%5Cbegin%7Bcases%7D%0Aa%3D5%5C%5C%0Ab%3D%5Cfrac%7B5%7D%7B2%7D%0A%5Cend%7Bcases%7D$ ，所以 $%5Cfrac%7Ba%7D%7Bb%7D%3D2$ .

10.

? $f%5Cleft(x%5Cright)%3Dx%5E2-2x-14%5Csqrt%7Bx-1%7D%2Bx%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%5Cleft(x%5Cgeq1%5Cright)$

的最小值為___________.

答案?? $-4$

解析??

$%5Cbegin%7Baligned%7D%0Af%5Cleft(x%5Cright)%26%3Dx%5E2-2x-14%5Csqrt%7Bx-1%7D%2Bx%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%5C%5C%26%0A%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft(x%2B%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%5Cright)%5E2-%5Cfrac%7B61%7D%7B2%7D%0A%5Cend%7Baligned%7D$

根據(jù)幾何意義

$%5Cbegin%7Baligned%7D%0Ax%2B%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%26%3D%5Csqrt%7B%5Cleft(x-2%5Cright)%5E2%2B%5Cleft(2%5Csqrt%7Bx-1%7D-0%5Cright)%5E2%7D%2B%5Csqrt%7B%5Cleft(x-4%5Cright)%5E2%2B%5Cleft(2%5Csqrt%7Bx-1%7D-7%5Cright)%5E2%7D%5C%5C%26%5Cgeq%5Csqrt%7B%5Cleft(4-2%5Cright)%5E2%2B%5Cleft(7-0%5Cright)%5E2%7D%5C%5C%26%3D%5Csqrt%7B53%7D%0A%5Cend%7Baligned%7D$

當(dāng)且僅當(dāng) $x%3D%5Cfrac%7B4%5Csqrt%7B53%7D%2B106%7D%7B49%7D$ 時取等.

所以

$f%5Cleft(x%5Cright)%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft(x%2B%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%5Cright)%5E2-%5Cfrac%7B61%7D%7B2%7D%5Cgeq-4$

當(dāng)且僅當(dāng) $x%3D%5Cfrac%7B4%5Csqrt%7B53%7D%2B106%7D%7B49%7D$ 時取等.

因此 $f%5Cleft(x%5Cright)$ 的最小值為 $-4$ .

11. 設(shè)復(fù)數(shù) $a$ ， $b$ ， $c$ 有 $%5Cleft%7Ca%5Cright%7C%5E2%2B%5Cleft%7Cb%5Cright%7C%5E2%2B%5Cleft%7Cc%5Cright%7C%5E2%3D1$ ，則

$%5Cleft%7Cab%5Cleft(a%5E2-b%5E2%5Cright)%2Bbc%5Cleft(b%5E2-c%5E2%5Cright)%2Bca%5Cleft(c%5E2-a%5E2%5Cright)%5Cright%7C$

的最大值為___________.

答案?? $%5Cfrac%7B9%7D%7B16%7D$

解析??

設(shè)

$a%3Dx_1%2By_1%5Cmathrm%7Bi%7D$

$b%3Dx_2%2By_2%5Cmathrm%7Bi%7D$

$c%3Dx_3%2By_3%5Cmathrm%7Bi%7D$

則

$%5Cbegin%7Baligned%7D%0A%26%5Cleft%7Cab%5Cleft(a%5E2-b%5E2%5Cright)%2Bbc%5Cleft(b%5E2-c%5E2%5Cright)%2Bca%5Cleft(c%5E2-a%5E2%5Cright)%5Cright%7C%5C%5C%26%3D%5Cleft%7C%5Cleft(a-b%5Cright)%5Cleft(b-c%5Cright)%5Cleft(c-a%5Cright)%5Cleft(a%2Bb%2Bc%5Cright)%5Cright%7C%5C%5C%26%0A%3D%5Cleft%7Ca-b%5Cright%7C%5Cleft%7Cb-c%5Cright%7C%5Cleft%7Cc-a%5Cright%7C%5Cleft%7Ca%2Bb%2Bc%5Cright%7C%5C%5C%26%0A%3D%5Csqrt%7B%5Cleft(x_1-x_2%5Cright)%5E2%2B%5Cleft(y_1-y_2%5Cright)%5E2%7D%5Ccdot%5Csqrt%7B%5Cleft(x_2-x_3%5Cright)%5E2%2B%5Cleft(y_2-y_3%5Cright)%5E2%7D%5Ccdot%5Csqrt%7B%5Cleft(x_3-x_1%5Cright)%5E2%2B%5Cleft(y_3-y_1%5Cright)%5E2%7D%20%5Ccdot%20%5Csqrt%7B%5Cleft(x_1%2Bx_2%2Bx_3%5Cright)%5E2%2B%5Cleft(y_1%2By_2%2By_3%5Cright)%5E2%7D%5C%5C%26%0A%5Cleq%5Cleft%5B%5Cfrac%7B%5Cleft(x_1-x_2%5Cright)%5E2%2B%5Cleft(y_1-y_2%5Cright)%5E2%2B%5Cleft(x_2-x_3%5Cright)%5E2%2B%5Cleft(y_2-y_3%5Cright)%5E2%2B%5Cleft(x_3-x_1%5Cright)%5E2%2B%5Cleft(y_3-y_1%5Cright)%5E2%2B%5Cleft(x_1%2Bx_2%2Bx_3%5Cright)%5E2%2B%5Cleft(y_1%2By_2%2By_3%5Cright)%5E2%7D%7B4%7D%5Cright%5D%5E2%5C%5C%26%0A%3D%5Cleft%5B%5Cfrac%7B3%5Cleft(x_1%5E2%2By_1%5E2%2Bx_2%5E2%2By_2%5E2%2Bx_3%5E2%2By_3%5E2%5Cright)%7D%7B4%7D%5Cright%5D%5E2%5C%5C%26%0A%3D%5Cleft%5B%5Cfrac%7B3%5Cleft(%5Cleft%7Ca%5Cright%7C%5E2%2B%5Cleft%7Cb%5Cright%7C%5E2%2B%5Cleft%7Cc%5Cright%7C%5E2%5Cright)%7D%7B4%7D%5Cright%5D%5E2%5C%5C%26%0A%3D%5Cfrac%7B9%7D%7B16%7D%0A%5Cend%7Baligned%7D$

當(dāng)且僅當(dāng) $a$ ， $b$ ， $c$ 對應(yīng)的以原點為起點的向量的終點構(gòu)成邊長為 $%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D$ ，重心到原點距離為 $%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B6%7D$ 的等邊三角形時取等.