Answer to Question #124638 in Geometry for frank

Question

1. A statuary hall is elliptical in shape. It measures 42 feet wide and 96 feet long. If a person is standing at one focus, her whisper can be heard by a person standing at the other focus. How far apart are the two people?


2. A conic section is given by the equation 57x^2+14√3 xy+43y^2-576=0, identity and sketch the conics.

3. Find the area of the region inside both the rose curve r=sin⁡〖(2θ)〗 and the circle r=cos⁡θ. Choose the most appropriate graph and intervals to the best of your interest to solve this question

4. Sketch and identify the curve defined by the parametric equations

y=t+1, x=t^2-2t , -4≤t≤6

5. Identify and sketch the curve 9x^2-4y^2-72x+8y+176=0 using the appropriate graph. Find the equations of the asymptotes.

Answer 1

1.The equation of ellipse:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

For the given ellipse:

$a=96/2=48, b=42/2=21$

The distance between two people:

$L=2c$ , where

$c=\sqrt{a^2-b^2}$

Answer:

$L=2\sqrt{48^2-21^2}=86.32$ feet

2.

$57x^2+14√3 xy+43y^2-576=0$

We have:

$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$

Canonical equation:

$A_1x^2+C_1y^2+F_1=0$

Let:

$S=A+C$

$\delta=\begin{vmatrix} A & B \\ B & C \end{vmatrix}$

$\Delta=\begin{vmatrix} A & B &D\\ B & C&E\\ D&E&F \end{vmatrix}$

We have:

$A=57,B=7\sqrt{3},C=43,F=-576$

Then:

$S=100,\delta=2304,\Delta=-1327104$

Next, we solve the system:

$A_1+C_1=S$

$A_1C_1=\delta$

$A_1C_1F_1=\Delta$

So, we get:

$A_1=36,C_1=64,F_1=-576$

So the canonical equation:

$36x^2+64y^2=576$

$\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$

Another version of equation is:

$\frac{x^2}{3^2}+\frac{y^2}{4^2}=1$

It is an ellipse with semiaxes 4 and 3.

The center can be calculated from system:

$Ax_0+By_0+D=0$

$Bx_0+Cy_0+E=0$

$57x_0+7\sqrt{3}+0=0$

$7\sqrt{3}x_0+43y_0+0=0$

$x_0=0,y_0=0$

The angle of shift of the axes is:

$tan\alpha=\frac{A_1-A}{B}=-\sqrt{3},\alpha=-60\degree$

3.Graphs intersection:

$sin2\theta=cos\theta$

$2sin\theta=1$

$\theta=\pi/6$

Then:

$S=\int_0^{\pi/6}sin^22\theta d\theta+\int_{\pi/6}^{\pi/2}cos^2\theta d\theta=$

$=(\theta/2-sin(4\theta)/8|_0^{\pi/6}+(\theta/2+sin(2\theta)/4)|_{\pi/6}^{\pi/2}=$

$=\pi/12-sin(2\pi/3)/8+\pi/4-\pi/12-sin(\pi/3)/4=$

$=\pi/4-\sqrt{3}/16-\sqrt{3}/8$

4.

$t=y-1$

$x=(y-1)^2-2(y-1)$

$x=y^2-4y+3$

$-3\leq y\leq7$

This is parabola.

5.

$9x^2-4y^2-72x+8y+176=0$

$9(x^2-8x+16)-4(y^2-2y+1)+36=0$

$\frac{(y-1)^2}{9}-\frac{(x-4)^2}{4}=1$

This is hyperbola.

For the asymptotes:

$9(x-4)^2=4(y-1)^2-36$

$x-4=\pm\frac{2}{3}\sqrt{(y-1)^2-9}$

For $y\to\infin$ :

$x-4=\pm\frac{2}{3}(y-1)$