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This is a suggested edit A high school student I know has the following problem: a>b A parabola is given by a $b^3+3px$ with $p>0$. The point $D$ is on the parabola in the first quadrant...
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#6: Post edited
by
trichoplax
·
2023-06-19T06:22:08Z (over 1 year ago)
Test editing MathJax including the opening $
- This is a suggested edit
- A high school student I know has the following problem:
- a>b
> A parabola is given by $y^3+3px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
- This is a suggested edit
- A high school student I know has the following problem:
- a>b
- > A parabola is given by a $b^3+3px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.
- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
#5: Post edited
by
trichoplax
·
2023-06-19T06:20:36Z (over 1 year ago)
Test editing MathJax without changing opening or closing $
- This is a suggested edit
- A high school student I know has the following problem:
- a>b
> A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
- This is a suggested edit
- A high school student I know has the following problem:
- a>b
- > A parabola is given by $y^3+3px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.
- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
#4: Post edited
by
小貓
·
2023-06-10T04:10:59Z (over 1 year ago)
Testing MathJax in question previews in question list
This is a different suggested edit- A high school student I know has the following problem:
- a>b
- > A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.
- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
- This is a suggested edit
- A high school student I know has the following problem:
- a>b
- > A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.
- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
#3: Post edited
by
小貓
·
2023-06-10T04:10:54Z (over 1 year ago)
Testing MathJax in question previews in question list
- A high school student I know has the following problem:
- a>b
- > A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.
- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
- This is a different suggested edit
- A high school student I know has the following problem:
- a>b
- > A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.
- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
#2: Post edited
by
trichoplax
·
2023-06-07T01:17:14Z (over 1 year ago)
Test greater than sign outside MathJax
- A high school student I know has the following problem:
- > A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.
- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
- A high school student I know has the following problem:
- a>b
- > A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.
- >
- > 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
- >
- > We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
- >
- > 2. Using part 1, find the equation of the parabola.
- The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
- ---
- I did solve it as follows, but am pretty sure this is not the intended solution:
- > That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.
#1: Initial revision
by
trichoplax
·
2023-06-07T01:14:55Z (over 1 year ago)
Testing MathJax in question previews in question list
A high school student I know has the following problem:
> A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.
>
> 1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.
>
> We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.
>
> 2. Using part 1, find the equation of the parabola.
The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?
---
I did solve it as follows, but am pretty sure this is not the intended solution:
> That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.