How to find the triangle area inside the parabola? Please help me understand.
$begingroup$
The parabola $C$ has cartesian equation $y^2 = 12x.$
The point $P(3p^2, 6p)$ lies on $C,$ where $pneq0.$
(a) Show that the equation of the normal to the curve $C$ at the point $P$ is
$$y + px = 6p + 3p^3$$
This normal crosses the curve $C$ again at the point $Q.$
Given that $p = 2$ and that $S$ is the focus of the parabola, find(b) the coordinates of the point $Q,$
(c) the area of the triangle $PQS.$
I can't figure out a way to solve question (c). I know the answer but don't understand it.
analytic-geometry conic-sections
edited Dec 31 '18 at 11:41
user376343
3,9584829
asked Dec 26 '18 at 20:07
AntonioAntonio
11
$endgroup$
add a comment |
$begingroup$
The parabola $C$ has cartesian equation $y^2 = 12x.$
The point $P(3p^2, 6p)$ lies on $C,$ where $pneq0.$
(a) Show that the equation of the normal to the curve $C$ at the point $P$ is
$$y + px = 6p + 3p^3$$
This normal crosses the curve $C$ again at the point $Q.$
Given that $p = 2$ and that $S$ is the focus of the parabola, find(b) the coordinates of the point $Q,$
(c) the area of the triangle $PQS.$
I can't figure out a way to solve question (c). I know the answer but don't understand it.
analytic-geometry conic-sections
edited Dec 31 '18 at 11:41
user376343
3,9584829
asked Dec 26 '18 at 20:07
AntonioAntonio
11
$endgroup$
$begingroup$
Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
$endgroup$
– Anurag A
Dec 26 '18 at 20:10
1
$begingroup$
Have you solved (a) and (b)? How did you do that? What did you get?
$endgroup$
– Arthur
Dec 26 '18 at 20:11
$begingroup$
Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
$endgroup$
– Antonio
Dec 26 '18 at 20:29
$begingroup$
If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
$endgroup$
– Aretino
Dec 28 '18 at 15:52
add a comment |
$begingroup$
The parabola $C$ has cartesian equation $y^2 = 12x.$
The point $P(3p^2, 6p)$ lies on $C,$ where $pneq0.$
(a) Show that the equation of the normal to the curve $C$ at the point $P$ is
$$y + px = 6p + 3p^3$$
This normal crosses the curve $C$ again at the point $Q.$
Given that $p = 2$ and that $S$ is the focus of the parabola, find(b) the coordinates of the point $Q,$
(c) the area of the triangle $PQS.$
I can't figure out a way to solve question (c). I know the answer but don't understand it.
analytic-geometry conic-sections
edited Dec 31 '18 at 11:41
user376343
3,9584829
asked Dec 26 '18 at 20:07
AntonioAntonio
11
$endgroup$
The parabola $C$ has cartesian equation $y^2 = 12x.$
The point $P(3p^2, 6p)$ lies on $C,$ where $pneq0.$
(a) Show that the equation of the normal to the curve $C$ at the point $P$ is
$$y + px = 6p + 3p^3$$
This normal crosses the curve $C$ again at the point $Q.$
Given that $p = 2$ and that $S$ is the focus of the parabola, find(b) the coordinates of the point $Q,$
(c) the area of the triangle $PQS.$
I can't figure out a way to solve question (c). I know the answer but don't understand it.
analytic-geometry conic-sections
analytic-geometry conic-sections
edited Dec 31 '18 at 11:41
user376343
3,9584829
asked Dec 26 '18 at 20:07
AntonioAntonio
11
edited Dec 31 '18 at 11:41
user376343
3,9584829
asked Dec 26 '18 at 20:07
AntonioAntonio
11
edited Dec 31 '18 at 11:41
user376343
3,9584829
edited Dec 31 '18 at 11:41
user376343
3,9584829
edited Dec 31 '18 at 11:41
user376343
3,9584829
3,9584829
asked Dec 26 '18 at 20:07
AntonioAntonio
11
asked Dec 26 '18 at 20:07
AntonioAntonio
11
asked Dec 26 '18 at 20:07
AntonioAntonio
11
11
$begingroup$
Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
$endgroup$
– Anurag A
Dec 26 '18 at 20:10
1
$begingroup$
Have you solved (a) and (b)? How did you do that? What did you get?
$endgroup$
– Arthur
Dec 26 '18 at 20:11
$begingroup$
Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
$endgroup$
– Antonio
Dec 26 '18 at 20:29
$begingroup$
If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
$endgroup$
– Aretino
Dec 28 '18 at 15:52
add a comment |
$begingroup$
Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
$endgroup$
– Anurag A
Dec 26 '18 at 20:10
1
$begingroup$
Have you solved (a) and (b)? How did you do that? What did you get?
$endgroup$
– Arthur
Dec 26 '18 at 20:11
$begingroup$
Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
$endgroup$
– Antonio
Dec 26 '18 at 20:29
$begingroup$
If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
$endgroup$
– Aretino
Dec 28 '18 at 15:52
$begingroup$
Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
$endgroup$
– Anurag A
Dec 26 '18 at 20:10
$begingroup$
Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
$endgroup$
– Anurag A
Dec 26 '18 at 20:10
1
1
$begingroup$
Have you solved (a) and (b)? How did you do that? What did you get?
$endgroup$
– Arthur
Dec 26 '18 at 20:11
$begingroup$
Have you solved (a) and (b)? How did you do that? What did you get?
$endgroup$
– Arthur
Dec 26 '18 at 20:11
$begingroup$
Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
$endgroup$
– Antonio
Dec 26 '18 at 20:29
$begingroup$
Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
$endgroup$
– Antonio
Dec 26 '18 at 20:29
$begingroup$
If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
$endgroup$
– Aretino
Dec 28 '18 at 15:52
$begingroup$
If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
$endgroup$
– Aretino
Dec 28 '18 at 15:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
$$y=-px+n$$ and plug in the coordinates of $P$ we get
$$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
And für $p=2$ you have to solve
$$(-2x+36)^2=12x$$ for $$x$$
answered Dec 26 '18 at 20:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
$begingroup$
And what about the triangle area PQS?
$endgroup$
– Antonio
Dec 26 '18 at 20:32
add a comment |
$begingroup$
For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
$$
area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
$$
answered Dec 29 '18 at 16:10
AretinoAretino
25.2k21445
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
$$y=-px+n$$ and plug in the coordinates of $P$ we get
$$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
And für $p=2$ you have to solve
$$(-2x+36)^2=12x$$ for $$x$$
answered Dec 26 '18 at 20:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
$begingroup$
And what about the triangle area PQS?
$endgroup$
– Antonio
Dec 26 '18 at 20:32
add a comment |
$begingroup$
The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
$$y=-px+n$$ and plug in the coordinates of $P$ we get
$$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
And für $p=2$ you have to solve
$$(-2x+36)^2=12x$$ for $$x$$
answered Dec 26 '18 at 20:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
$begingroup$
And what about the triangle area PQS?
$endgroup$
– Antonio
Dec 26 '18 at 20:32
add a comment |
$begingroup$
The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
$$y=-px+n$$ and plug in the coordinates of $P$ we get
$$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
And für $p=2$ you have to solve
$$(-2x+36)^2=12x$$ for $$x$$
answered Dec 26 '18 at 20:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
$$y=-px+n$$ and plug in the coordinates of $P$ we get
$$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
And für $p=2$ you have to solve
$$(-2x+36)^2=12x$$ for $$x$$
answered Dec 26 '18 at 20:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Dec 26 '18 at 20:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Dec 26 '18 at 20:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Dec 26 '18 at 20:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
77.9k42866
$begingroup$
And what about the triangle area PQS?
$endgroup$
– Antonio
Dec 26 '18 at 20:32
add a comment |
$begingroup$
And what about the triangle area PQS?
$endgroup$
– Antonio
Dec 26 '18 at 20:32
$begingroup$
And what about the triangle area PQS?
$endgroup$
– Antonio
Dec 26 '18 at 20:32
$begingroup$
And what about the triangle area PQS?
$endgroup$
– Antonio
Dec 26 '18 at 20:32
add a comment |
$begingroup$
For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
$$
area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
$$
answered Dec 29 '18 at 16:10
AretinoAretino
25.2k21445
$endgroup$
add a comment |
$begingroup$
For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
$$
area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
$$
answered Dec 29 '18 at 16:10
AretinoAretino
25.2k21445
$endgroup$
add a comment |
$begingroup$
For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
$$
area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
$$
answered Dec 29 '18 at 16:10
AretinoAretino
25.2k21445
$endgroup$
For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
$$
area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
$$
answered Dec 29 '18 at 16:10
AretinoAretino
25.2k21445
answered Dec 29 '18 at 16:10
AretinoAretino
25.2k21445
answered Dec 29 '18 at 16:10
AretinoAretino
25.2k21445
answered Dec 29 '18 at 16:10
AretinoAretino
25.2k21445
25.2k21445
add a comment |
add a comment |
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$begingroup$
Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
$endgroup$
– Anurag A
Dec 26 '18 at 20:10
1
$begingroup$
Have you solved (a) and (b)? How did you do that? What did you get?
$endgroup$
– Arthur
Dec 26 '18 at 20:11
$begingroup$
Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
$endgroup$
– Antonio
Dec 26 '18 at 20:29
$begingroup$
If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
$endgroup$
– Aretino
Dec 28 '18 at 15:52