About Lesson
5.3 Ellipse 椭圆
1. Find the centre, vertex, axis of symmetry, major semi-axis, minor semi-axis, eccentricity, focus, directrix and length of latus rectum for the following ellipse and sketch the graph.
求以下各椭圆的中心, 顶点, 对称轴, 半长轴之长, 半短轴之长, 离心率, 焦点, 准线和通径长度並描其图形
(a)
(b)
(c)
(d)
(e)
Answer (a) :
Answer (b) :
Answer (c) :
Answer (d) :
Answer (e) :
2. Find the equation of the ellipse with center at (-3, 1), one of vertices at (-5, 1), length of latus rectum equal to 1 and
求椭圆方程式其中心在(-3, 1), 其中一顶点在(-5, 1), 通径长度为1且
求椭圆方程式其中心在(-3, 1), 其中一顶点在(-5, 1), 通径长度为1且
(a) major axis parallel x-axis,
长轴平行于x轴,
(b) major axis parallel y-axis,
长轴平行于y轴,
Answer :
(a)
![Rendered by QuickLaTeX.com (x + 3)^2 + 4(y - 1)^2 = 4](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-01085f254cede23d0d5b81a578b25f5b_l3.png)
(b)
3. Find the equation of the ellipse which has same eccentricity and left directrix with ellipse
, and also take its right focus to be the left focus.
![Rendered by QuickLaTeX.com x^2 + 2y^2 = 2](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-3c2c005b83a3058d23b9788d98feb197_l3.png)
求与椭圆有相同的离心率和公共的左准线,且以它的右焦点为左焦点的椭圆方程.
Answer :
![Rendered by QuickLaTeX.com \frac{{(x-4)}^2}{18}+\frac{y^2}{9}=1](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-29efc7d24cd3b5e4ba6cb86dfe027f69_l3.png)
4. Given that the major axis of an ellipse is 4, y-axis be its directrix, the left focus is on the parabola
. Find the equation of the ellipse when eccentricity is
.
![Rendered by QuickLaTeX.com y^2 = x - 1](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-d68e0a4a795cb6c2809e76640e4bb8e4_l3.png)
![Rendered by QuickLaTeX.com \frac{2}{3}](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-63e3d8a8d7f540b1a64ce735a01e9b38_l3.png)
己知椭圆的长轴长为4,以y轴为准线,左焦点在拋物线上。求当离心率为
时, 此椭圆方程式。
Answer :
![Rendered by QuickLaTeX.com \frac{{(x-3)}^2}{4}+\frac{9{(y\pm\sqrt{2/3})}^2}{20}=100](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-767869bd054668b112ce4dafdc99353b_l3.png)
5. Given that F
(2, -2), F
(2, 0) are the foci of a ellipse, straight line y = 3 is a directrix of the ellipse, find the equation of the ellipse.
![Rendered by QuickLaTeX.com _1](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-3a8732cc72efffee8872fbe29964e4a0_l3.png)
![Rendered by QuickLaTeX.com _2](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-5e62d6979a14842f0671d1eb51126a78_l3.png)
已知椭圆的焦点为F(2, -2), F
(2, 0), 直线y = 3是椭圆的一条准线, 求椭圆的方程.
Answer :
![Rendered by QuickLaTeX.com \frac{{(x-2)}^2}{3}+\frac{{(y+1)}^2}{4}=1](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-5b7e42944c1d385453f558b3655721a3_l3.png)
6. Find the equation of the ellipse with vertices at (0, -1) and (12, -1), a focus at
.
![Rendered by QuickLaTeX.com (6 + \sqrt{11}, -1)](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-ed5a530a2350d1e40d11d8eb2344a64a_l3.png)
求顶点在 (0, -1) 和 (12, -1), 一焦点在 的椭圆方程式
Answer :
![Rendered by QuickLaTeX.com \frac{{(x-6)}^2}{36}+\frac{{(y+1)}^2}{25}=1](https://learn-ondemand.com/wp-content/ql-cache/quicklatex.com-9b5cf36e336873e4edf390c7606a3a6e_l3.png)