Hi everyone! I'm taking a math course in the summer( not for a credit just to learn) and I seriously don't get how to do these questions. Can someone please help.
1) A bubble over a set of tennis courts is semi elliptical in cross section, 20m wide at the ground and 8m high in the center. How high is the bubble at a point 2m in from the outer edge. Answer: 4.8m
2) Halley's comet appears to move around the sun, one focus of its elliptical orbit, every 76 years. The major axis of this orbit is 5270 million km and the eccentricity of the orbit is 0.97, where eccentricity is e=c/a
a) find an equation for the orbit
B) find the min distance btw the comet and the sun(the perigee of the orbit)
Answer: 80.5 million km.

2. Originally Posted by swtaznjackie
Hi everyone! I'm taking a math course in the summer( not for a credit just to learn) and I seriously don't get how to do these questions. Can someone please help.
1) A bubble over a set of tennis courts is semi elliptical in cross section, 20m wide at the ground and 8m high in the center. How high is the bubble at a point 2m in from the outer edge. Answer: 4.8m
The cross section is a semi-circle of radius $\displaystyle 8$m, stretched by a factor of $\displaystyle 2.5$
in the horizontal direction.a point $\displaystyle 2$m from the outer edge corresponds to a
point $\displaystyle 2/2.5$ m from the outer edge of the $\displaystyle 8$ m radius circle, which we may take
to have origin at it's centre and so equation:

$\displaystyle x^2+y^2=64$

and the $\displaystyle x$ coordinate of the point in question id $\displaystyle 8-2/2.5=18/2.5$, so the
height is:

$\displaystyle y=\sqrt{64-(18/2.5)^2}~= 3.49$ m

RonL

3. Originally Posted by swtaznjackie
2) Halley's comet appears to move around the sun, one focus of its elliptical orbit, every 76 years. The major axis of this orbit is 5270 million km and the eccentricity of the orbit is 0.97, where eccentricity is e=c/a
a) find an equation for the orbit
B) find the min distance btw the comet and the sun(the perigee of the orbit)
Answer: 80.5 million km.
a) As $\displaystyle \varepsilon=0.97$ we have $\displaystyle \varepsilon=\sqrt{1-b^2/a^2}$ so $\displaystyle b=a\sqrt{1-\varepsilon^2}\approx 640.58$ km.

Taking the centre of the ellipse as origin we have:

$\displaystyle \frac{x^2}{2635^2}+\frac{y^2}{640.58^2} = 1$

RonL

4. Originally Posted by swtaznjackie
2) Halley's comet appears to move around the sun, one focus of its elliptical orbit, every 76 years. The major axis of this orbit is 5270 million km and the eccentricity of the orbit is 0.97, where eccentricity is e=c/a
a) find an equation for the orbit
B) find the min distance btw the comet and the sun(the perigee of the orbit)
Answer: 80.5 million km.
b) We also have $\displaystyle \varepsilon=c/a$ where $\displaystyle c$ is the distance of a focus from the centre of the ellipse, so $\displaystyle c=\varepsilon \times a \approx 2555.95$ km, so the perihelion is $\displaystyle a-c=5270/2-2555.95=79.05$ km

RonL

5. Thanks everyone for the help!

6. Originally Posted by swtaznjackie
1) A bubble over a set of tennis courts is semi elliptical in cross section, 20m wide at the ground and 8m high in the center. How high is the bubble at a point 2m in from the outer edge. Answer: 4.8m
A standard form of a hrizontal ellipse is
[(x-h)/a]^2 +[(y-k)/b]^2 = 1
where
a = semi-major axis---->20/2 = 10m here.
b = semi-minor axis----> 8m here.
(h,k) is the center of the ellipse.

Let's put the (h,k) at (0,0), so, the equation of the ellipse is
(x/10)^2 +(y/8)^2 = 1
(x^2)/100 +(y^2)/64 = 1
Clear the fractions,multiply both sides by 100(64),
64x^2 +100y^2 = 100(64) ------------(i)

Now, the point 2m from an end is (10-2) = 8m from the center. Or, the said point is at x=8m. What is the y at x=8?
64(8^2) +100y^2 = 100(64)
100y^2 = 100(64) -64(64)
100y^2 = 64(100-64)
100y^2 = 64(36)
Take the squareroot of both sides,
10y = 8(6)
y = 48/10 = 4.8m --------------answer.