Tutors Answer Your Questions about Quadratic-relations-and-conic-sections (FREE)
Question 1028938: Using the formula y= -2/225x^2 +50 (for outputs greater than 0.)to determine the shape of a satellite dish what are the minimal interior dimensions of a shipping crate needed for the satellite dish if the machine that is making the dish with that formula is programmed in cm? The only information given is the formula used by the dish-making machine.
Click here to see answer by josmiceli(19441)   |
Question 1028982: For each equation of the parabola, (a) reduce to standard and then find the (b) direction of opening, (c) vertex, (d) focus, (e) endpoints of the latus rectum, and (d) equation of directrix. Graph the parabola.
x^2 2x 36y + 1 = 0
(x - 〖3)〗^2 = -12 (y 4 )
y^2 + 24x + 48 = 0
Click here to see answer by josgarithmetic(39618) |
Question 1028984: Write the general and standard form of the equation of the circle satisfying the given conditions.
With center (2, -1) and radius 3
With center (-4, 5) and passing through (2, 1)
Having the points (-1, -1) and ( 3, 4 ) as ends of the diameter
Center at ( 4, 1 ) and touching the y- axis
With center at (-2, -1 ) and tangent to the line 4x-3y = 12
Passing through ( 1,2), (2,3) and (-2, 1)
Passing through (4, 6), (-2,-2) and (-4, 2)
Tangent to the line 3x- 4y - 5 = 0 at (3,1) and passing through ( -3, -1)
Inscribed in a triangle with sides on the lines x 3y = -5, 3x + y = 1 and 3x y = -11.
Find the general equation of the tangent to the circle
(x - 〖3)〗^2 + (y - 〖5)〗^2 = 64 at point ( 3, 3 )
x^2 + y^2 2x 24 = 0 at point ( -2, -4 )
x^2 + y^2 2x 24 = 0 at point ( -2, -4 )
x^2 + y^2 6x 4y - 28 = 0 parallel to the line 4x + 5y = 8
x^2 + y^2 4y 9 = 0 through ( 5 , -1 )
x^2 + y^2 6x 2y + 5 = 0 through ( -3, 6 )
For each pair of equations of circles, find the general equation of the radical axis
x^2 + y^2 14y + 40 = 0 and x^2 + y^2 = 4
x^2 + y^2 14x 12y + 65 = 0 and x^2 + y^2 6x 4y + 3 = 0
x^2 + y^2 12x + 14y + 60 = 0 and x^2 + y^2 + 6x + 4y - 3 = 0
Click here to see answer by Alan3354(69443)  |
Question 1029039: I was absent the day we studied this and the teacher won't help me out- could you explain how I would do a problem like this?
Use the focus-directrix definition of a parabola to answer the following question
1) How would the shape of the parabola change if the focus were moved up, away from the directrix? How would we describe p?
Click here to see answer by josgarithmetic(39618) |
Question 1029075: Could someone please check my work on this? I apologize for the length, but would really appreciate some help! "The shape of the Gateway Arch can be approx modeled with the equations of 2 parabolas: one for the outer/upper surface, and one for the inner/lower surface. The height of the arch is 630 ft, and at ground level the outsides of the bases are 630 ft apart. The arch narrows as it rises. Therefore, the insides of the bases are only 540 feet apart at ground level, but the inside of the arch has a height of 615 ft. suppose you are standing at the origin which is at ground level directly underneath the center of the arch.
a) find the vertex and x-intercepts for both the outer and inner parabolas.
**My answer*** Outer parabola vertex: (0,630), x-intercepts (-315,0)(315,0) and inner parabola vertex: (0,615), x-intercepts: (-270,0) (270,0). Is this correct???
b) find the height above the ground of the focus and directrix for each parabola. Round your answer to the nearest tenth of a foot.
***my answer** outer parabola focus height: 590.625 and directrix height: 1220.625 feet.
Inner parabola focus height: 585.37 ft and directrix height: 1200.37 ft.
c) Find the equation for both outer and inner parabolas in standard form
**my answer** Inner: y=-(41/4860)x^2 + 615 and outer: y=-(2/315)x^2+630. I'm not really sure if this is even standard form???
I appreciate any help and again, I'm sorry it's such a long and involved question!
Click here to see answer by Fombitz(32388)  |
Question 1030334: I've been trying to solve and understand these problems for the past 2 hours, and it's killing me. I still don't understand it. Can you please show the work when solving, in hopes that I can see, and understand.
4y^2-16x+8y-12=0
I need to rewrite in standard form, as well as find the vertex,focus, and directrix.
Click here to see answer by rothauserc(4718)  |
Question 1031364: Consider the graph y=(x-k)^2 where k is any interger. What effect does changing the value of have on:
The axis of symmetry
The turning point
The x and y intercepts
The shape of the curve
You must present your findings for each specific value of k and generalise the the effects in the terms of K.
Thanks.
Click here to see answer by Edwin McCravy(20056)  |
Question 1031395: Are these 3 points the vertices of a right triangle (-9,5), (-7,9), (-5,8).
I answered no but not sure if I am doing it right.
Also are these 3 points Collinear (-7,0), (-2,4), (-11,-5)
I answered no on that one also.
Can you explain how you got your answer so I know how to do it please.
Click here to see answer by robertb(5830)  |
Question 1031488: I'm really struggling with these types of problems.
Find the equation of the parabola with the given information, then find the two points that define the latus rectum.
Focus(-5,-4) and directrix x =3.
I think what I mostly need help with is finding the vertex so I can use that info to find the equation. I'm not sure I understand how to find the vertex. Thank you.
Click here to see answer by Fombitz(32388) |
Question 1032842: Find the center, foci, vertices, asymptotes, and radius, as appropriate,
of the conic section:
-------------------------------------------
Here is what I did...I am just not sure if I did it correctly.
=
=
=
=
=
=
=
=
Center: (7, -3)
Foci: (7,  - 3)
Vertices: (7, -2)
Asymptotes: N/A
Radius: N/A
Did I do this right? I just have the gut feeling that I messed up somewhere.
Thank you!
Click here to see answer by Alan3354(69443) |
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