
Tutors Answer Your Questions about Quadraticrelationsandconicsections (FREE)
Question 994389: Use the following equation: y = –3(x + 8)^2 + 2
The coordinates of the vertex written as (h, k) would be?
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! They are (8,2). You reverse the sign of the constant in the parentheses and take the last constant without changing sign.
This is also
f(x)=3x^248x190.
The vertex is b/2a which is 48/6 =8
f(x)=192+384190=2
Question 993785: for the parabola : y = x^2 + 36.
Graph of a parabola opening down at the vertex (0,36) crossing the x–axis at (6,0) and (6,0).
Answer by solver91311(20879) (Show Source):
Question 993571: find the equation of a circle that passes thru'(6,0) and (24,0) and is tangent to the yaxis
Found 2 solutions by solver91311, anand429: Answer by solver91311(20879) (Show Source):
You can put this solution on YOUR website!
The segment of the xaxis bounded by the points (6,0) and (24,0) is a chord of the desired circle. Since the perpendicular bisector of a chord must pass through the center of the circle, we know that the center of the desired circle lies on the line x = 15. We also know, since said line is a bisector of the chord, that the distance from the point (6,0) to the point of intersection of the line x = 15 and the xaxis is 9. Since the circle is tangent to the yaxis, the distance from the yaxis to the center of the circle must be 15, hence the radius of the circle is 15.
Consider the right triangle formed by the radius of the circle that has the point (6,0) as an end point, the portion of the line x = 15 from the center of the circle to the center of the chord segment of the xaxis, and the halfchord from point (6,0) to (15,0). Since this is a right triangle with hypotenuse/short leg ratio of 5:3, the other leg must be in ratio 5:4:3 with the other two sides. Hence the distance from the xaxis to the center of the circle is 12.
Our circle is centered at (15,12) and has a radius of 15.
A circle centered at (h, k) with radius r has an equation:
John
My calculator said it, I believe it, that settles it
Answer by anand429(129) (Show Source):
You can put this solution on YOUR website! (6,0) and (24,0) both lie on xaxis, hence mid point of these points will lie directly below( or above) centre (Since line joining centre to mid point of a chord is perpendicular to the chord)
Also, since the circle touches the y axis, so distance from origin to this mid point found above will be equal to radius.
Now mid point is ((6+24)/2,(0+0)/2) i.e. (15,0)
So radius is 15.
Now let distance of centre from mid point found above be p
So, (DRAW DIAGRAM TO UNDERSTAND ALL THE STEPS)
i.e.
=> p = 12 or p = 12
So coordinates of centre will become (15,12) or (15,12)
So, the equation of circle,
or
=> or
(DRAW DIAGRAM TO UNDERSTAND ALL THE STEPS)
Question 993227: find the standard form of the equation of the parabola satisfying the given conditions
focus (2,1) directrix x=4
Answer by josgarithmetic(13975) (Show Source):
Question 992413: Equation of parabola and latus rectum with focus (8,0) and directrix x=8
Answer by josgarithmetic(13975) (Show Source):
Question 992302: Hyperbola
Determine the center,vertices, foci,conjugat, and the latus rectum of the
(y+3)^2/16  (x2)^2/25=1
Answer by MathLover1(11324) (Show Source):
You can put this solution on YOUR website! ....if you compare to you see that ,, ,
since the part of the equation is added, then the center, foci, and vertices will be above and below the center (on a line paralleling the yaxis)
so,
the center is at (,)= (,)
semimajor axis length
semiminor axis length
since and , the equation tells me that , so , and
the eccentricity is
the vertices and foci are above and below the center,
so the foci are at
(,)=(,) and
(,)= (,)
or, approximately at (, ) and (,)
the vertices are at (,) and (, )
(,) =>(,)
and
(, )=>(,)
The length of the Latus Rectum:
In a hyperbola, it is twice the square of the length of the transverse axis divided by the length of the conjugate axis.
the length of the transverse axis is =>=>
the length of the conjugate axis is =>=>
the Latus Rectum is
Question 992315: directrix has yintercept 3
Answer by ikleyn(988) (Show Source):
Question 992151: x^2/4+y^2/9+z^2/16=1
Answer by ikleyn(988) (Show Source):
Question 992106: Find the equation of the parabola with focus (5,4) and directrix x = 3
Answer by josgarithmetic(13975) (Show Source):
Question 991825: Please help me with this! Find the centre, vertices, lenght of latus rectum, foci , eccentricities of the ellipse x^2 + 9y^2 + 4x 18y  23 = 0
Answer by ikleyn(988) (Show Source):
Question 991566: How do you the parabola y=4x^212x+5 into vertex form and intercept form?
Answer by Fombitz(25151) (Show Source):
Question 991544: What is the vertex form of the equation y=4(x+1)(x1)?
Answer by Fombitz(25151) (Show Source):
Question 991186: 4x^2+9y^2=25 at (2,1)
Answer by Alan3354(47455) (Show Source):
Question 990195: I need help Graphing y=x^23x+6. Can someone please help me.
Answer by josgarithmetic(13975) (Show Source):
Question 988025: Question Part
Points
Submissions Used
Use the quadratic formula to find any xintercepts of the parabola. (If an answer does not exist, enter DNE.)
y = 2x^2 − 10x + 5
Answer by josmiceli(13716) (Show Source):
Question 987973: (Ellipse)
25x^2 + 64y^2  400x  600y=0
Find its standard equation, foci, vertices and extemeties of the minor axis.
Answer by MathLover1(11324) (Show Source):
You can put this solution on YOUR website!
.........complete square
.........since ...
....multiply by
....divide by
now we have a standard form and we know that:
=>
=>
we can calculate using formula =>
we also know that
=>so, the center is at (, ), approximately at (, )
foci:(, ) or (, ), approximately at (, ) or (, )
vertices: (, ) or (, ), approximately at (, ) or (, )
semimajor axis length: ~~
semiminor axis length: (~~
eccentricity  ~~
Question 987549: Test for symmetry
4x^2=+y^2=64
Answer by Alan3354(47455) (Show Source):
Question 987347: what are the coordinates of the center of the ellipse graphed by the equation below? (x+14)^2/6^2+(y+17)^2/5^2=1?
Answer by macston(4006) (Show Source):
Question 986810: Put the equation in standard form for a hyperbola.
4x2−9y2−24x−54y=81
Answer by josgarithmetic(13975) (Show Source):
Question 986493: Sketch the graph of 9x2 – 16y2 = 144. Label any intercepts and asymptotes.
Answer by Fombitz(25151) (Show Source):
Question 986364: How do I algebraically determine where these two conic sections intersect?
4x^2+16y^2=64
2xy^2=4
Answer by Alan3354(47455) (Show Source):
Question 986363: How can I algebraically find out where these two conic sections intersect?
4x^2+16y^2=64
2xy^2=4
Answer by Alan3354(47455) (Show Source):
You can put this solution on YOUR website! How can I algebraically find out where these two conic sections intersect?
4x^2+16y^2=64
2xy^2=4
=====================
I would solve the 2nd eqn for x, then sub for x in the 1st eqn.
Question 985894: How do you find the focus and directrix of the parabola y=x^24x+4 and graph it?
Answer by MathLover1(11324) (Show Source):
Question 985669: a.
What is the maximum value of this expression?
b.
For which values of y will x be positive?
Please show working, thanks in advance
Found 2 solutions by Alan3354, macston: Answer by Alan3354(47455) (Show Source): Answer by macston(4006) (Show Source):
Question 985675:
For which values of x will y be negative?
Answer by Alan3354(47455) (Show Source):
You can put this solution on YOUR website!
For which values of x will y be negative?
================
Find the 2 zeroes:
y = (x1)*(x5)
y = 0 at x = 1 and at x = 5
y < 0 for 1 < x < 5
Question 985667:
What is the smallest positive integer for b that will force this parabola to have two xintercepts?
Please show working
Answer by Edwin McCravy(13211) (Show Source):
Question 985482: For the circle and the line , determine the exact values of the gradient so that the line:
a) is a tangent to the circle
b) intersects the circle in two places
c) does not intersect the circle
Found 3 solutions by ikleyn, solver91311, KMST: Answer by ikleyn(988) (Show Source):
You can put this solution on YOUR website!
To find the intersection points of the circle and the straight line, substitute = into the equation = . You will get
+ = ,
+ = ,
= ,
= .
Next, apply the quadratic formula to find the roots
= ,
where is the discriminant = = = .
Now everything is determined by the discriminant.
If > , then there are two roots. Correspondingly, there are two intersection points.
In opposite, if there are two intersection points, then there are two roots, hence, > .
If = , then there is only one root. It means that the straight line tangents the circle.
And in opposite, if the straight line tangents the circle, then there is only one root, hence, = .
Finally, < <> there are no intersection points.
Thus you need to solve this critical equation
= .
It gives
= +/ .
So, if = +/ , then the straight line tangents the circle.
If > , then > and there are two intersection points.
If < < , then < and there are no intersection points.
Answer by solver91311(20879) (Show Source):
You can put this solution on YOUR website!
KMSTs solution is more general, and therefore better, IMHO, than what I am about to show you. However, this is a slightly different way to look at this particular problem and may give you some additional insight.
Note that a tangent to a circle is perforce perpendicular to a radius. An examination of the diagram shows us that a tangent as described, the perpendicular radius, and that segment of the axis between the origin and the intercept of the desired line form a right triangle. Since the measure of the radius of the circle which forms the short leg of the right triangle is 2 and the measure of the axis segment that forms the hypotenuse of the right triangle measures 4, we have a 306090 right triangle, where the angle between the radius and the axis is the 60 degree angle. See figure below:
.
That means that the angle between the radius ray and the axis is 30 degrees. A quick examination of a unit circle and using a multiplier of 2 to match the radius of the given circle, we see readily that the point of tangency in QI is .
Using the slope formula, the slope of the line that passes through the points and is .
Symmetry gives us a second point of tangency in QII, which is simply and then the slope of the line through the second point is just the additive inverse of the slope of the first equation.
From there it should be clear that any slope in the open interval will produce a secant line.
Further, any slope in either open interval or will produce a line that does not intersect the circle.
John
My calculator said it, I believe it, that settles it
Answer by KMST(3791) (Show Source):
You can put this solution on YOUR website! gives us the intersection points of circle and line.
>>>
That is a quadratic equation.
A quadratic equation of the form
has no real solutions if .
It has one real solution if , and
it has two real solutions if .
For the quadratic equation , ,so
.
The zeros of are the zeros of , given by
and .
For those values, the line is tangent to the circle.
In between those values of ,
for ,
is negative, and the line does not intersect the circle.
Otherwise, for ,
, and the line intersects the circle in two places.
Question 984896: Please help me answer this equation:
Identify the vertex, focus,and directrix of equation:
y=1/12x^2
Found 2 solutions by macston, josgarithmetic: Answer by macston(4006) (Show Source): Answer by josgarithmetic(13975) (Show Source):
Question 984794: Find the fifth roots of 32(cos 280° + i sin 280°).
Answer by srinivas.g(527) (Show Source):
You can put this solution on YOUR website! i) Let z = 32(cos 280° + i sin 280°) = 32{cos(k*360° + 280°) + isin(k*360° + 280°)},
where k is an integer.
ii) ==> z^(1/5) = [32{cos(k*360° + 280°) + isin(k*360° + 280°)}]^(1/5)
==> z^(1/5) = 2[cos{(k*360° + 280°)/5} + isin{(k*360° + 280°)/5}]
[Application of DeMoievere's theorem]
==> z^(1/5) = 2[cos(72k + 56) + isin(72k + 56)]
We can now get the 5 roots, by assigning k = 0, 1, 2, 3 & 4
When k = 0, 1st root = 2(cos 56° + isin 56°)
When k = 1, 2nd root = 2(cos 128° + isin 128°)
When k = 2, 3rd root = 2(cos 200° + isin200°)
When k = 3, 4th root = 2(cos 272° + isin 272°)
When k = 4, 5th root = 2(cos 344° + isin344°)
Question 984603: Find an equation of the parabola described then graph the parabola. Focus(2,1); Vertex(3,1)
Answer by josgarithmetic(13975) (Show Source):
Question 984535: I have two limit questions:
Find the derivative of f(x) = 2/x at x = 2.
Find the derivative of f(x) = 9/x at x = 8.
Please help!
Answer by jim_thompson5910(33401) (Show Source):
Question 984414: solve the system.
x+y2z=8
x+3y+2z=14
x3y2z=4
Answer by Fombitz(25151) (Show Source):
Question 984409: I have two limit questions:
Find the derivative of f(x) = 5/x at x = 1.
Use graphs and tables to find the limit and identify any vertical asymptotes of lim(x>7) 1/(x7)^2
Answer by solver91311(20879) (Show Source):
Question 983852: What is the standard equation of the parabola with given: vertices (2,5) parallel to xaxis and passes through (7,1)
Answer by Timnewman(249) (Show Source):
Question 983937: The height of the water, H, in feet, at a boat dock t hours after 6 a.m. is given by H=8+sin pi/3 t. Find the height of the water at 9a.m.
Answer by solver91311(20879) (Show Source):
You can put this solution on YOUR website!
9 AM is three hours after 6 AM. Plug in 3 for t and do the arithmetic.
John
My calculator said it, I believe it, that settles it
Question 983860: What is the standard equation of the parabola with given vertices (2,5), parallel axis to x axis and if its passes through (7,1)
Answer by josgarithmetic(13975) (Show Source):
Question 983479: Write the equation of an ellipse with vertices at (7, 0) and (7, 0) and covertices at (0, 1) and (0, 1).
Answer by KMST(3791) (Show Source):
Question 983482: Write the equation of an ellipse with a major axis of length 8 and covertices (0, 3) and (0, 3).
Answer by josgarithmetic(13975) (Show Source):
Question 983483: Write the equation of an ellipse with vertices (0, 12) and (0, 12) and covertices (2, 0) and (2, 0).
Answer by josgarithmetic(13975) (Show Source):
Question 983481: What is the equation of a parabola with a directrix x = 8 and focus (8, 0).
Answer by josgarithmetic(13975) (Show Source):

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