How might I quickly determine the equation of the parabola, given the coordinates of its focus and vertex? The 2019 Stack Overflow Developer Survey Results Are InEquation of a parabola: Translations and directrixesHow to derive the equation of a parabola given a focus and a directrix not parallel to the x or y axis?Focus And Vertex Of An Inclined ParabolaFind the equation of the parabola with focus (2;1) and vertex in the origin.Find Vertex when Focus and Directrix of Parabola is given.What's the difference between these two approaches to find the equation of a parabola?Equivalent definition of a parabola as a locus of points given its focus and vertexHow to find the coefficients of a parabola if only its focus and directrix are knownFind the equation of parabola, when tangent at two points and vertex is givenFocus of parabola with directrix $x=2$ and vertex not at origin?
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How might I quickly determine the equation of the parabola, given the coordinates of its focus and vertex?
The 2019 Stack Overflow Developer Survey Results Are InEquation of a parabola: Translations and directrixesHow to derive the equation of a parabola given a focus and a directrix not parallel to the x or y axis?Focus And Vertex Of An Inclined ParabolaFind the equation of the parabola with focus (2;1) and vertex in the origin.Find Vertex when Focus and Directrix of Parabola is given.What's the difference between these two approaches to find the equation of a parabola?Equivalent definition of a parabola as a locus of points given its focus and vertexHow to find the coefficients of a parabola if only its focus and directrix are knownFind the equation of parabola, when tangent at two points and vertex is givenFocus of parabola with directrix $x=2$ and vertex not at origin?
$begingroup$
I have this MCQ.
Which of the following is the equation of the parabola with focus at
$(1,2)$ and vertex at
$(3,2)$ :
A. $ y^2 - 4y + 8x - 20 = 0 $
B. $ y^2 + y + 8x -20 = 0 $
C. $ y^2 + 4y + 8x - 20 = 0 $
D. $ y^2 - 4y + x - 20 = 0 $
E. $ y^2 - 4y + 8x + 20 = 0 $
Now, I know how to find equation of a parabola from focus and vertex, or from directrix and vertex, but that takes some time. I was wondering if there is a way to just sort of see which equation is the right one, because right from the focus and vertex we can see that the parabola will open on the left and its equation will be $y^2 = - 4ax$.
I'm not sure if it's actually possible though, but I think it would be cool if there was a way.
geometry conic-sections
edited Mar 23 at 11:48
Aretino
25.8k31545
asked Mar 22 at 23:56
ArkiloArkilo
555
$endgroup$
|
show 1 more comment
$begingroup$
I have this MCQ.
Which of the following is the equation of the parabola with focus at
$(1,2)$ and vertex at
$(3,2)$ :
A. $ y^2 - 4y + 8x - 20 = 0 $
B. $ y^2 + y + 8x -20 = 0 $
C. $ y^2 + 4y + 8x - 20 = 0 $
D. $ y^2 - 4y + x - 20 = 0 $
E. $ y^2 - 4y + 8x + 20 = 0 $
Now, I know how to find equation of a parabola from focus and vertex, or from directrix and vertex, but that takes some time. I was wondering if there is a way to just sort of see which equation is the right one, because right from the focus and vertex we can see that the parabola will open on the left and its equation will be $y^2 = - 4ax$.
I'm not sure if it's actually possible though, but I think it would be cool if there was a way.
geometry conic-sections
edited Mar 23 at 11:48
Aretino
25.8k31545
asked Mar 22 at 23:56
ArkiloArkilo
555
$endgroup$
3
$begingroup$
If you know how to do the work then do the work.
$endgroup$
– John Douma
Mar 23 at 0:03
1
$begingroup$
For a parabola like $y=ax^2+bx+c$ the vertex is located at the point where the derivative vanishes. This is, when $2ax+b=0$, or what is the same $x=-fracb2a$. In your case $x$ and $y$ are exchanged. You can see that only A,D, and E result in a vertex where $y=2$. Of those only A results in $x=3$ when $y=2$. Then you only need to check if A has focus at $(1,2)$.
$endgroup$
– user647486
Mar 23 at 0:11
2
$begingroup$
Why are people voting this question down? This genuinely perplexes me!
$endgroup$
– Theo Bendit
Mar 23 at 0:13
1
$begingroup$
By the way, the equation wouldn't be $y^2 = -4ax$, it would be $(y-colorredk)^2 = -4a(x-colorredh)$, where $(h,k)$ is the vertex (which is known here). Also, $a$ is the focal length, which you can deduce since you are given the focus and vertex. So since you can easily work out $a$ and know $h,k$, you can put them into $(y-colorredk)^2 = -4a(x-colorredh)$, and then expand and rearrange to get the desired answer.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:18
1
$begingroup$
The vertex tells you that A. is the only possibility.
$endgroup$
– herb steinberg
Mar 23 at 0:38
|
show 1 more comment
$begingroup$
I have this MCQ.
Which of the following is the equation of the parabola with focus at
$(1,2)$ and vertex at
$(3,2)$ :
A. $ y^2 - 4y + 8x - 20 = 0 $
B. $ y^2 + y + 8x -20 = 0 $
C. $ y^2 + 4y + 8x - 20 = 0 $
D. $ y^2 - 4y + x - 20 = 0 $
E. $ y^2 - 4y + 8x + 20 = 0 $
Now, I know how to find equation of a parabola from focus and vertex, or from directrix and vertex, but that takes some time. I was wondering if there is a way to just sort of see which equation is the right one, because right from the focus and vertex we can see that the parabola will open on the left and its equation will be $y^2 = - 4ax$.
I'm not sure if it's actually possible though, but I think it would be cool if there was a way.
geometry conic-sections
edited Mar 23 at 11:48
Aretino
25.8k31545
asked Mar 22 at 23:56
ArkiloArkilo
555
$endgroup$
I have this MCQ.
Which of the following is the equation of the parabola with focus at
$(1,2)$ and vertex at
$(3,2)$ :
A. $ y^2 - 4y + 8x - 20 = 0 $
B. $ y^2 + y + 8x -20 = 0 $
C. $ y^2 + 4y + 8x - 20 = 0 $
D. $ y^2 - 4y + x - 20 = 0 $
E. $ y^2 - 4y + 8x + 20 = 0 $
Now, I know how to find equation of a parabola from focus and vertex, or from directrix and vertex, but that takes some time. I was wondering if there is a way to just sort of see which equation is the right one, because right from the focus and vertex we can see that the parabola will open on the left and its equation will be $y^2 = - 4ax$.
I'm not sure if it's actually possible though, but I think it would be cool if there was a way.
geometry conic-sections
geometry conic-sections
edited Mar 23 at 11:48
Aretino
25.8k31545
asked Mar 22 at 23:56
ArkiloArkilo
555
edited Mar 23 at 11:48
Aretino
25.8k31545
asked Mar 22 at 23:56
ArkiloArkilo
555
edited Mar 23 at 11:48
Aretino
25.8k31545
edited Mar 23 at 11:48
Aretino
25.8k31545
edited Mar 23 at 11:48
Aretino
25.8k31545
25.8k31545
asked Mar 22 at 23:56
ArkiloArkilo
555
asked Mar 22 at 23:56
ArkiloArkilo
555
asked Mar 22 at 23:56
ArkiloArkilo
555
555
3
$begingroup$
If you know how to do the work then do the work.
$endgroup$
– John Douma
Mar 23 at 0:03
1
$begingroup$
For a parabola like $y=ax^2+bx+c$ the vertex is located at the point where the derivative vanishes. This is, when $2ax+b=0$, or what is the same $x=-fracb2a$. In your case $x$ and $y$ are exchanged. You can see that only A,D, and E result in a vertex where $y=2$. Of those only A results in $x=3$ when $y=2$. Then you only need to check if A has focus at $(1,2)$.
$endgroup$
– user647486
Mar 23 at 0:11
2
$begingroup$
Why are people voting this question down? This genuinely perplexes me!
$endgroup$
– Theo Bendit
Mar 23 at 0:13
1
$begingroup$
By the way, the equation wouldn't be $y^2 = -4ax$, it would be $(y-colorredk)^2 = -4a(x-colorredh)$, where $(h,k)$ is the vertex (which is known here). Also, $a$ is the focal length, which you can deduce since you are given the focus and vertex. So since you can easily work out $a$ and know $h,k$, you can put them into $(y-colorredk)^2 = -4a(x-colorredh)$, and then expand and rearrange to get the desired answer.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:18
1
$begingroup$
The vertex tells you that A. is the only possibility.
$endgroup$
– herb steinberg
Mar 23 at 0:38
|
show 1 more comment
3
$begingroup$
If you know how to do the work then do the work.
$endgroup$
– John Douma
Mar 23 at 0:03
1
$begingroup$
For a parabola like $y=ax^2+bx+c$ the vertex is located at the point where the derivative vanishes. This is, when $2ax+b=0$, or what is the same $x=-fracb2a$. In your case $x$ and $y$ are exchanged. You can see that only A,D, and E result in a vertex where $y=2$. Of those only A results in $x=3$ when $y=2$. Then you only need to check if A has focus at $(1,2)$.
$endgroup$
– user647486
Mar 23 at 0:11
2
$begingroup$
Why are people voting this question down? This genuinely perplexes me!
$endgroup$
– Theo Bendit
Mar 23 at 0:13
1
$begingroup$
By the way, the equation wouldn't be $y^2 = -4ax$, it would be $(y-colorredk)^2 = -4a(x-colorredh)$, where $(h,k)$ is the vertex (which is known here). Also, $a$ is the focal length, which you can deduce since you are given the focus and vertex. So since you can easily work out $a$ and know $h,k$, you can put them into $(y-colorredk)^2 = -4a(x-colorredh)$, and then expand and rearrange to get the desired answer.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:18
1
$begingroup$
The vertex tells you that A. is the only possibility.
$endgroup$
– herb steinberg
Mar 23 at 0:38
3
3
$begingroup$
If you know how to do the work then do the work.
$endgroup$
– John Douma
Mar 23 at 0:03
$begingroup$
If you know how to do the work then do the work.
$endgroup$
– John Douma
Mar 23 at 0:03
1
1
$begingroup$
For a parabola like $y=ax^2+bx+c$ the vertex is located at the point where the derivative vanishes. This is, when $2ax+b=0$, or what is the same $x=-fracb2a$. In your case $x$ and $y$ are exchanged. You can see that only A,D, and E result in a vertex where $y=2$. Of those only A results in $x=3$ when $y=2$. Then you only need to check if A has focus at $(1,2)$.
$endgroup$
– user647486
Mar 23 at 0:11
$begingroup$
For a parabola like $y=ax^2+bx+c$ the vertex is located at the point where the derivative vanishes. This is, when $2ax+b=0$, or what is the same $x=-fracb2a$. In your case $x$ and $y$ are exchanged. You can see that only A,D, and E result in a vertex where $y=2$. Of those only A results in $x=3$ when $y=2$. Then you only need to check if A has focus at $(1,2)$.
$endgroup$
– user647486
Mar 23 at 0:11
2
2
$begingroup$
Why are people voting this question down? This genuinely perplexes me!
$endgroup$
– Theo Bendit
Mar 23 at 0:13
$begingroup$
Why are people voting this question down? This genuinely perplexes me!
$endgroup$
– Theo Bendit
Mar 23 at 0:13
1
1
$begingroup$
By the way, the equation wouldn't be $y^2 = -4ax$, it would be $(y-colorredk)^2 = -4a(x-colorredh)$, where $(h,k)$ is the vertex (which is known here). Also, $a$ is the focal length, which you can deduce since you are given the focus and vertex. So since you can easily work out $a$ and know $h,k$, you can put them into $(y-colorredk)^2 = -4a(x-colorredh)$, and then expand and rearrange to get the desired answer.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:18
$begingroup$
By the way, the equation wouldn't be $y^2 = -4ax$, it would be $(y-colorredk)^2 = -4a(x-colorredh)$, where $(h,k)$ is the vertex (which is known here). Also, $a$ is the focal length, which you can deduce since you are given the focus and vertex. So since you can easily work out $a$ and know $h,k$, you can put them into $(y-colorredk)^2 = -4a(x-colorredh)$, and then expand and rearrange to get the desired answer.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:18
1
1
$begingroup$
The vertex tells you that A. is the only possibility.
$endgroup$
– herb steinberg
Mar 23 at 0:38
$begingroup$
The vertex tells you that A. is the only possibility.
$endgroup$
– herb steinberg
Mar 23 at 0:38
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
You don't need any particular formula here: just plug the coordinates of the vertex into the equation and you'll see that only in case A the point belongs to the parabola.
answered Mar 23 at 11:51
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
Jesus how did I forget to PITA
$endgroup$
– Arkilo
Mar 23 at 13:14
add a comment |
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$begingroup$
You don't need any particular formula here: just plug the coordinates of the vertex into the equation and you'll see that only in case A the point belongs to the parabola.
answered Mar 23 at 11:51
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
Jesus how did I forget to PITA
$endgroup$
– Arkilo
Mar 23 at 13:14
add a comment |
$begingroup$
You don't need any particular formula here: just plug the coordinates of the vertex into the equation and you'll see that only in case A the point belongs to the parabola.
answered Mar 23 at 11:51
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
Jesus how did I forget to PITA
$endgroup$
– Arkilo
Mar 23 at 13:14
add a comment |
$begingroup$
You don't need any particular formula here: just plug the coordinates of the vertex into the equation and you'll see that only in case A the point belongs to the parabola.
answered Mar 23 at 11:51
AretinoAretino
25.8k31545
$endgroup$
You don't need any particular formula here: just plug the coordinates of the vertex into the equation and you'll see that only in case A the point belongs to the parabola.
answered Mar 23 at 11:51
AretinoAretino
25.8k31545
answered Mar 23 at 11:51
AretinoAretino
25.8k31545
answered Mar 23 at 11:51
AretinoAretino
25.8k31545
answered Mar 23 at 11:51
AretinoAretino
25.8k31545
25.8k31545
$begingroup$
Jesus how did I forget to PITA
$endgroup$
– Arkilo
Mar 23 at 13:14
add a comment |
$begingroup$
Jesus how did I forget to PITA
$endgroup$
– Arkilo
Mar 23 at 13:14
$begingroup$
Jesus how did I forget to PITA
$endgroup$
– Arkilo
Mar 23 at 13:14
$begingroup$
Jesus how did I forget to PITA
$endgroup$
– Arkilo
Mar 23 at 13:14
add a comment |
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3
$begingroup$
If you know how to do the work then do the work.
$endgroup$
– John Douma
Mar 23 at 0:03
1
$begingroup$
For a parabola like $y=ax^2+bx+c$ the vertex is located at the point where the derivative vanishes. This is, when $2ax+b=0$, or what is the same $x=-fracb2a$. In your case $x$ and $y$ are exchanged. You can see that only A,D, and E result in a vertex where $y=2$. Of those only A results in $x=3$ when $y=2$. Then you only need to check if A has focus at $(1,2)$.
$endgroup$
– user647486
Mar 23 at 0:11
2
$begingroup$
Why are people voting this question down? This genuinely perplexes me!
$endgroup$
– Theo Bendit
Mar 23 at 0:13
1
$begingroup$
By the way, the equation wouldn't be $y^2 = -4ax$, it would be $(y-colorredk)^2 = -4a(x-colorredh)$, where $(h,k)$ is the vertex (which is known here). Also, $a$ is the focal length, which you can deduce since you are given the focus and vertex. So since you can easily work out $a$ and know $h,k$, you can put them into $(y-colorredk)^2 = -4a(x-colorredh)$, and then expand and rearrange to get the desired answer.
$endgroup$
– Minus One-Twelfth
Mar 23 at 0:18
1
$begingroup$
The vertex tells you that A. is the only possibility.
$endgroup$
– herb steinberg
Mar 23 at 0:38