# Thread: Conics-Hyperbolas, Circle, Parabola Exam Question

1. ## Conics-Hyperbolas, Circle, Parabola Exam Question

Greeting I am facing difficulties with certain questions that my teacher said I would need to know for my test which on Wednesday and I was wondering if I can get some help. Any help would be greatly appreciated, if I am correct these are old exam questions, but i am really having a tough time solving these questions. Any help would be greatly appreciated.

2. Originally Posted by Solid8Snake
Greeting I am facing difficulties with certain questions that my teacher said I would need to know for my test which on Wednesday and I was wondering if I can get some help. Any help would be greatly appreciated, if I am correct these are old exam questions, but i am really having a tough time solving these questions. Any help would be greatly appreciated.
to #3:

1. The origin is the center of the hyperbola. Thus the general equation of the hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

2. You know at least 2 points of the hyperbola: (2,0), (4,2). The coordinates of these points must satisfy the equation of the hyperbola. Plug in the coordinates and solve the system of equations for a and b. I've got $a=2~\wedge~b=\sqrt{\dfrac43}$

The equation of the hyperbola is: $\dfrac{x^2}{4}-\dfrac{y^2}{\frac43}=1$

3. The points B and D are symmetric by reflection over the x-axis. You only need to know the y-coordinate of B to calculate the distance BD. The x-coordinate of B is 3. Plug in this value and solve the equation of the hyperbola for y:

$\dfrac{9}{4}-\dfrac{y^2}{\frac43}=1~\implies~y^2=\dfrac53$

The point B has the coordinates: $\left(3,\ \sqrt{\dfrac53}\right)$; point D is at $\left(3,\ -\sqrt{\dfrac53}\right)$

4. $|\overline{BD}|=2 \cdot \sqrt{\dfrac53} \approx 2.582\ cm$ rounded to the nearest centimeter the distance is 3 cm.

3. Originally Posted by Solid8Snake
Greeting I am facing difficulties with certain questions that my teacher said I would need to know for my test which on Wednesday and I was wondering if I can get some help. Any help would be greatly appreciated, if I am correct these are old exam questions, but i am really having a tough time solving these questions. Any help would be greatly appreciated.
to #2:

1. The radius of the circle is perpendicular to the tangent in point T(9, 4). The slope of the radius is: $m_r=\dfrac{4-6}{9-8}=-2$. Therefore the slope of the tangent is $m_t=\dfrac12$

2. Use the slope-point-formula of a straight line to get the equation of the tangent:

$y=\dfrac12 x - \dfrac12$

3. The corner of the room lies on the tangent at x = 8. Therefore the corner has the coordinates C(8, 3.5)

the lowest point of the circle has the coordinates $L(8, 6-\sqrt{5})$

4. The distance between the corner and the disk is:

$d=|\overline{CL}| = 8-\sqrt{5}-3.5\approx 0.2639$. Rounded as requested

$d \approx 0.3$

4. Thx for the help, however number one is still very confusing it was the hardest out of all of them I find if you can help me with that that would be great. Thx for your help concerning 2 and 3 I understand it perfectly now.

5. Originally Posted by Solid8Snake
Greeting I am facing difficulties with certain questions that my teacher said I would need to know for my test which on Wednesday and I was wondering if I can get some help. Any help would be greatly appreciated, if I am correct these are old exam questions, but i am really having a tough time solving these questions. Any help would be greatly appreciated.
to #1:

I've divided the height of the trophy into 3 different parts: The top distance t in red, the middle distance m in brown and the bottom b distance in blue. (See attachment).

1. b is the y-value of the hyperbola if x = 10:

$\dfrac{100}4-\dfrac{y^2}{16}=1~\implies~y=-\sqrt{16 \cdot 24}\approx -19.5959$

Therefore $\boxed{b = 19.5959}$

2. m is the y-value of the hyperbola if x = 3.5:

$\dfrac{12.25}4-\dfrac{y^2}{16}=1~\implies~y=\sqrt{33}\approx 5.7446$

Therefore $\boxed{m = 5.7446}$

3. To calculate t I placed the center of the ellipse on the origin. The semi-axes of the ellipse are A = 4 and B = 8. Thus the equation of this ellipse is:

$\dfrac{x^2}{16}+\dfrac{y^2}{64}=1$

Then t = B + |y(3.5)| (That means that the top distance consists of the semi-axis B (= 8) and the y-value if x = 3.5):

$\dfrac{12.25}{16}+\dfrac{y^2}{64}=1~\implies~y=\pm \sqrt{15}\approx 3.8730$

Therefore the top distance is: $\boxed{t = 8 + 3.8730 = 11.8730}$

4. The total height of the trophy is the sum of the three different distances: $\boxed{\bold{H = 27.2135}}$

6. Thx for the help, but it appears your response to number one is wrong the answer i got was 50.13 about and the teacher said the answer should be close to 49 I guess my answer is close enough, thx for the help though it helped me determine how to solve the question.