Lecture 19
2.2.2 Circles
A circle centered at the origin $(0,0)$ with radius $r$ is given by the equation
$ x^2 + y^2 = r^2.$
2.2.2 Circles
A circle with radius $r$ and center $(h,k)$ corresponds to
$(x-h)^2 + (y-k)^2 = r^2.$
2.2.2 Circles
Example: Sketch the circle $x^2+10x+y^2-4y+20=0.$
Complete the square:
$ \left( \left(x+5\right)^2- 25\right)$ $ + \left( \left(y-2\right)^2-4\right)$ $ +\,20=0$
$\Ra \left(x+5\right)^2+ \left(y-2\right)^2 =9$ $ = 3^2$
Circle of radius 3, centre at $(-5,2)$
Skecht it! 📝
Just touching the heart by Inigo Quilez
2.2.3 Ellipses
In standard form, an ellipse centered at the origin is given by the equation
$\ds \frac{x^2}{a^2} + \frac{y^2}{b^2} =1.$
Let $a\gt b\,$ and define $c\geq 0\, $ by $c^2=a^2-b^2$
$y=0\,$ $\Ra x= a,-a$
$x=0\,$ $\Ra y= \pm b$
An ellipse may also be defined as the set of points in a plane the sum of whose distances from two fixed focal points $F_1$ and $F_2$ is a constant, and you should definitely ask your lecturer to use two pins and a piece of string to demonstrate this.
An ellipse may also be defined as the set of points in a plane the sum of whose distances from two fixed focal points $F_1$ and $F_2$ is a constant, and you should definitely ask your lecturer to use two pins and a piece of string to demonstrate this.
An ellipse may also be defined as the set of points in a plane the sum of whose distances from two fixed focal points $F_1$ and $F_2$ is a constant, and you should definitely ask your lecturer to use two pins and a piece of string to demonstrate this.
If $a \geq b,$ then the foci are at $(\pm c,0)$ where $c^2 = a^2 -b^2.$ These lie on the major axis ($x$-axis here). The ellipse intersects the $x$ axis at the vertices $(\pm a,0),$ and the $y$-axis at the vertices $(0,\pm b).$
2.2.3 Ellipses
Question: What is the equation of an ellipse centered at the point $(h,k)$?
$\ds \frac{ \left( x-h\right)^2 }{a^2}+ \frac{ \left( y-k\right)^2 }{b^2}=1$
2.2.3 Ellipses
Example: Sketch the ellipse $x^2-x+9y^2=0$.
Complete the square:
$ \ds \left( \left(x-\frac{1}{2}\right)^2- \frac{1}{4}\right)$ $ +\, 9y^2$ $ =0$
$\Ra \ds \left(x-\frac{1}{2}\right)^2+ 9y^2 =\frac{1}{4}$
$\Ra \ds 4\left(x-\frac{1}{2}\right)^2+ 36y^2 =1$
2.2.3 Ellipses
Example: Sketch the ellipse $x^2-x+9y^2=0$.
$ \ds \left( \left(x-\frac{1}{2}\right)^2- \frac{1}{4}\right)$ $ +\, 9y^2$ $ =0$
$\Ra \ds 4\left(x-\frac{1}{2}\right)^2+ 36y^2 =1$
$\Ra \ds \frac{\left(x-\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2}+ \frac{y^2}{\left(\frac{1}{6}\right)^2} =1$
2.2.3 Ellipses
Example: Sketch the ellipse $x^2-x+9y^2=0$.
$\Ra \ds \frac{\left(x-\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2}+ \frac{y^2}{\left(\frac{1}{6}\right)^2} =1$
If $\, y=0\,$ $\, \ds \Ra x-\frac{1}{2} = \pm \frac{1}{2}\,$ $\, \ds \Ra x = \frac{1}{2}\pm \frac{1}{2}\,$
If $\,\ds x=\frac{1}{2}\,$ $\,\ds \Ra y = \pm \frac{1}{6}$
2.2.3 Ellipses
Example: Sketch the ellipse $x^2-x+9y^2=0$.
$\Ra \ds \frac{\left(x-\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2}+ \frac{y^2}{\left(\frac{1}{6}\right)^2} =1$
If $\, y=0\,$ $\, \ds \Ra x = \frac{1}{2}\pm \frac{1}{2}\,$
If $\,\ds x=\frac{1}{2}\,$ $\,\ds \Ra y = \pm \frac{1}{6}$
2.2.4 Hyperbolas
A hyperbola centered at the origin, with asymptotes $y=\pm b x/a,$ is given by the equation
$\ds \frac{x^2}{a^2} - \frac{y^2}{b^2} =1 \;\;$ or by $\;\;\ds-\frac{x^2}{a^2} + \frac{y^2}{b^2} =1.$
A hyperbola may also be defined as the set of points in a plane, the difference of whose distances from two fixed focal points $F_1$ and $F_2$ is a constant.
A hyperbola may also be defined as the set of points in a plane, the difference of whose distances from two fixed focal points $F_1$ and $F_2$ is a constant.
For such a hyperbola, the foci are at $(\pm c,0)$ and $(0,\pm c)$ respectively, where $c^2 = a^2 + b^2.$
2.2.4 Hyperbolas
Question: What is the equation of a hyperbola centered at the point $(h,k)$?
$\ds \frac{ \left( x-h\right)^2 }{a^2}- \frac{ \left( y-k\right)^2 }{b^2}=\pm1$
2.2.4 Hyperbolas
Example: Sketch the hyperbola $x^2-4y^2=9$.
Standard form:
$ \ds \frac{x^2}{3^2} - \frac{2^2y^2}{3^2}=1$ $\,\Ra \, \ds \frac{x^2}{3^2} - \frac{y^2}{\left(\frac{3}{2}\right)^2}=1$
Centre at $\,(0,0)\,$ and $\,a = 3, b = \dfrac{3}{2}.$
Asymptotes are $\ds y =\pm \frac{\frac{3}{2}}{3}x$ $\ds = \pm \frac{3}{6}x$ $\ds = \pm \frac{1}{2}x$
Skecht it! 📝
a) $\,x^2-y^2-1=0\quad$ | d) $\,x^2-2x-y+1=0\quad\quad$ |
b) $\,x+y^2-1=0$ | e) $\,x^2-y^2+4=0$. |
c) $\,x^2-2x+y^2=0\quad\quad$ | f) $\,4x^2-8x+y^2=0$. |
2.2.6 Main points
Geographical maps have curves of constant height above sea level, or curves of constant air pressure (isobars), or curves of constant temperature (isothermals). Drawing contours is an effective method of representing a 3-dimensional surface in two dimensions.