Area of a triangle inside an ellipse
up vote
4
down vote
favorite
$F_1$, $F_2$ are are foci of the ellipse $dfrac{x^2}{9}+dfrac{y^2}{4}=1$.
$P$ is a point on the ellipse such that $|PF_1|:|PF_2|=2:1;$, then how could I figure out of the area of $∆PF_1F_2$?
As we know $c^2=a^2-b^2=9-4=5$,
$$∴c=pm sqrt5$$ So the epicenter of the ellipse respectively $(sqrt5,0); & ;(-sqrt5,0)$.
We've to find the area of $∆PF_1F_2$ which is $=1/2times F_1F_2times$(perpendicular distance from $P$ to any point of the horizontal line $F_1F_2$)
I'm not understanding what & how to do next... find out the area of ∆PF₁F₂
calculus triangle conic-sections area
asked Dec 4 at 5:49
RA FI
213
add a comment |
up vote
4
down vote
favorite
$F_1$, $F_2$ are are foci of the ellipse $dfrac{x^2}{9}+dfrac{y^2}{4}=1$.
$P$ is a point on the ellipse such that $|PF_1|:|PF_2|=2:1;$, then how could I figure out of the area of $∆PF_1F_2$?
As we know $c^2=a^2-b^2=9-4=5$,
$$∴c=pm sqrt5$$ So the epicenter of the ellipse respectively $(sqrt5,0); & ;(-sqrt5,0)$.
We've to find the area of $∆PF_1F_2$ which is $=1/2times F_1F_2times$(perpendicular distance from $P$ to any point of the horizontal line $F_1F_2$)
I'm not understanding what & how to do next... find out the area of ∆PF₁F₂
calculus triangle conic-sections area
asked Dec 4 at 5:49
RA FI
213
2
Hint: $|PF_1|^2 + |PF_2|^2 = 4^2 + 2^2 = 20 = |F_1F_2|^2$
– achille hui
Dec 4 at 8:56
1
guys I didn't get the point... I don't understanding anyone's hint😔.By the way this is my first question.. So I'm really happy for you guys commenting in my question ❤
– RA FI
Dec 4 at 9:23
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
$F_1$, $F_2$ are are foci of the ellipse $dfrac{x^2}{9}+dfrac{y^2}{4}=1$.
$P$ is a point on the ellipse such that $|PF_1|:|PF_2|=2:1;$, then how could I figure out of the area of $∆PF_1F_2$?
As we know $c^2=a^2-b^2=9-4=5$,
$$∴c=pm sqrt5$$ So the epicenter of the ellipse respectively $(sqrt5,0); & ;(-sqrt5,0)$.
We've to find the area of $∆PF_1F_2$ which is $=1/2times F_1F_2times$(perpendicular distance from $P$ to any point of the horizontal line $F_1F_2$)
I'm not understanding what & how to do next... find out the area of ∆PF₁F₂
calculus triangle conic-sections area
asked Dec 4 at 5:49
RA FI
213
$F_1$, $F_2$ are are foci of the ellipse $dfrac{x^2}{9}+dfrac{y^2}{4}=1$.
$P$ is a point on the ellipse such that $|PF_1|:|PF_2|=2:1;$, then how could I figure out of the area of $∆PF_1F_2$?
As we know $c^2=a^2-b^2=9-4=5$,
$$∴c=pm sqrt5$$ So the epicenter of the ellipse respectively $(sqrt5,0); & ;(-sqrt5,0)$.
We've to find the area of $∆PF_1F_2$ which is $=1/2times F_1F_2times$(perpendicular distance from $P$ to any point of the horizontal line $F_1F_2$)
I'm not understanding what & how to do next... find out the area of ∆PF₁F₂
calculus triangle conic-sections area
calculus triangle conic-sections area
asked Dec 4 at 5:49
RA FI
213
asked Dec 4 at 5:49
RA FI
213
edited Dec 4 at 14:46
asked Dec 4 at 5:49
RA FI
213
asked Dec 4 at 5:49
RA FI
213
asked Dec 4 at 5:49
RA FI
213
213
2
Hint: $|PF_1|^2 + |PF_2|^2 = 4^2 + 2^2 = 20 = |F_1F_2|^2$
– achille hui
Dec 4 at 8:56
1
guys I didn't get the point... I don't understanding anyone's hint😔.By the way this is my first question.. So I'm really happy for you guys commenting in my question ❤
– RA FI
Dec 4 at 9:23
add a comment |
2
Hint: $|PF_1|^2 + |PF_2|^2 = 4^2 + 2^2 = 20 = |F_1F_2|^2$
– achille hui
Dec 4 at 8:56
1
guys I didn't get the point... I don't understanding anyone's hint😔.By the way this is my first question.. So I'm really happy for you guys commenting in my question ❤
– RA FI
Dec 4 at 9:23
2
2
Hint: $|PF_1|^2 + |PF_2|^2 = 4^2 + 2^2 = 20 = |F_1F_2|^2$
– achille hui
Dec 4 at 8:56
Hint: $|PF_1|^2 + |PF_2|^2 = 4^2 + 2^2 = 20 = |F_1F_2|^2$
– achille hui
Dec 4 at 8:56
1
1
guys I didn't get the point... I don't understanding anyone's hint😔.By the way this is my first question.. So I'm really happy for you guys commenting in my question ❤
– RA FI
Dec 4 at 9:23
guys I didn't get the point... I don't understanding anyone's hint😔.By the way this is my first question.. So I'm really happy for you guys commenting in my question ❤
– RA FI
Dec 4 at 9:23
add a comment |
3 Answers
3
active
oldest
votes
up vote
5
down vote
Referring to the graph:
$hspace{2cm}$
Note that $|PF_1|+|PF_2|=|AF_1|+|AF_2|=2a=6$. Using the given condition $|PF_1|=2|PF_2|$, we find $|PF_1|=4, |PF_2|=2$. You can use Heron's formula to find the area of $Delta PF_1F_2$:
$$ a=|PF_1|=4, b=|PF_2|=2, c=|F_1F_2|=2sqrt{5}, p=frac{a+b+c}{2}=3+sqrt{5}; \
S=sqrt{p(p-a)(p-b)(p-c)}=sqrt{(3+sqrt{5})(sqrt{5}-1)(sqrt{5}+1)(3-sqrt{5})}=4.$$
answered Dec 4 at 10:31
farruhota
18.6k2736
I don't know how should I give you thanks...your answer helps me very much.... ❤❤❤
– RA FI
Dec 4 at 10:57
You are welcome. Note that achille hui hinted to the Pythagoras theorem, implying the triange was right. Good luck.
– farruhota
Dec 4 at 10:59
How could you find out the graph?any mobile apps?
– RA FI
Dec 4 at 11:07
I used desmos for raw graph and edited it.
– farruhota
Dec 4 at 11:09
You can also use graphing calculator.By the way how do u edit this? can I do this in mobile?
– RA FI
Dec 4 at 14:33
|
show 1 more comment
up vote
5
down vote
With the use of the definition of ellipse $$|PF_1|+|PF_2|=2a=6$$ and the given ratio $$|PF_1|:|PF_2|=2:1,$$ we get $$|PF_1|=4, ; |PF_2|=2$$
Since $F_1F_2=2sqrt 5=sqrt{20}=sqrt{4^2+2^2},$ we have a right triangle $triangle PF_1F_2$ with hypotenuse $F_1F_2.$ The area is $$mathcal{A}=frac 12 cdot |PF_1|cdot |PF_2|=4$$
answered Dec 4 at 10:40
user376343
2,7412822
1
Oh you've done with very simple method. thanks 😍❤
– RA FI
Dec 4 at 11:03
The school exercises often consider special (even trivial) cases :) Oh, I am reading comments and answers only now and I see there are traces indicating the same...
– user376343
Dec 4 at 11:09
Are u talking about the emojis?😅😂
– RA FI
Dec 4 at 14:21
My phone keyboard has this emojis.Are you using from computer?
– RA FI
Dec 4 at 14:52
Thanks, I didn't think this could be used within MSE 🦊
– user376343
Dec 4 at 15:13
add a comment |
up vote
3
down vote
Hint:
WLOG $P(3cos t,2sin t)$
$$|PF_1|^2=(3cos t-sqrt5)^2+(2sin t-0)^2=?$$
$$|PF_2|^2=(3cos t+sqrt5)^2+(2sin t-0)^2=?$$
$$dfrac{|PF_1|^2}{|PF_2|^2}=2^2$$
Use $sin^2t=1-cos^2t$ to form a Quadratic Equation in $cos t$
answered Dec 4 at 6:00
lab bhattacharjee
222k15155273
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Referring to the graph:
$hspace{2cm}$
Note that $|PF_1|+|PF_2|=|AF_1|+|AF_2|=2a=6$. Using the given condition $|PF_1|=2|PF_2|$, we find $|PF_1|=4, |PF_2|=2$. You can use Heron's formula to find the area of $Delta PF_1F_2$:
$$ a=|PF_1|=4, b=|PF_2|=2, c=|F_1F_2|=2sqrt{5}, p=frac{a+b+c}{2}=3+sqrt{5}; \
S=sqrt{p(p-a)(p-b)(p-c)}=sqrt{(3+sqrt{5})(sqrt{5}-1)(sqrt{5}+1)(3-sqrt{5})}=4.$$
answered Dec 4 at 10:31
farruhota
18.6k2736
I don't know how should I give you thanks...your answer helps me very much.... ❤❤❤
– RA FI
Dec 4 at 10:57
You are welcome. Note that achille hui hinted to the Pythagoras theorem, implying the triange was right. Good luck.
– farruhota
Dec 4 at 10:59
How could you find out the graph?any mobile apps?
– RA FI
Dec 4 at 11:07
I used desmos for raw graph and edited it.
– farruhota
Dec 4 at 11:09
You can also use graphing calculator.By the way how do u edit this? can I do this in mobile?
– RA FI
Dec 4 at 14:33
|
show 1 more comment
up vote
5
down vote
Referring to the graph:
$hspace{2cm}$
Note that $|PF_1|+|PF_2|=|AF_1|+|AF_2|=2a=6$. Using the given condition $|PF_1|=2|PF_2|$, we find $|PF_1|=4, |PF_2|=2$. You can use Heron's formula to find the area of $Delta PF_1F_2$:
$$ a=|PF_1|=4, b=|PF_2|=2, c=|F_1F_2|=2sqrt{5}, p=frac{a+b+c}{2}=3+sqrt{5}; \
S=sqrt{p(p-a)(p-b)(p-c)}=sqrt{(3+sqrt{5})(sqrt{5}-1)(sqrt{5}+1)(3-sqrt{5})}=4.$$
answered Dec 4 at 10:31
farruhota
18.6k2736
I don't know how should I give you thanks...your answer helps me very much.... ❤❤❤
– RA FI
Dec 4 at 10:57
You are welcome. Note that achille hui hinted to the Pythagoras theorem, implying the triange was right. Good luck.
– farruhota
Dec 4 at 10:59
How could you find out the graph?any mobile apps?
– RA FI
Dec 4 at 11:07
I used desmos for raw graph and edited it.
– farruhota
Dec 4 at 11:09
You can also use graphing calculator.By the way how do u edit this? can I do this in mobile?
– RA FI
Dec 4 at 14:33
|
show 1 more comment
up vote
5
down vote
up vote
5
down vote
Referring to the graph:
$hspace{2cm}$
Note that $|PF_1|+|PF_2|=|AF_1|+|AF_2|=2a=6$. Using the given condition $|PF_1|=2|PF_2|$, we find $|PF_1|=4, |PF_2|=2$. You can use Heron's formula to find the area of $Delta PF_1F_2$:
$$ a=|PF_1|=4, b=|PF_2|=2, c=|F_1F_2|=2sqrt{5}, p=frac{a+b+c}{2}=3+sqrt{5}; \
S=sqrt{p(p-a)(p-b)(p-c)}=sqrt{(3+sqrt{5})(sqrt{5}-1)(sqrt{5}+1)(3-sqrt{5})}=4.$$
answered Dec 4 at 10:31
farruhota
18.6k2736
Referring to the graph:
$hspace{2cm}$
Note that $|PF_1|+|PF_2|=|AF_1|+|AF_2|=2a=6$. Using the given condition $|PF_1|=2|PF_2|$, we find $|PF_1|=4, |PF_2|=2$. You can use Heron's formula to find the area of $Delta PF_1F_2$:
$$ a=|PF_1|=4, b=|PF_2|=2, c=|F_1F_2|=2sqrt{5}, p=frac{a+b+c}{2}=3+sqrt{5}; \
S=sqrt{p(p-a)(p-b)(p-c)}=sqrt{(3+sqrt{5})(sqrt{5}-1)(sqrt{5}+1)(3-sqrt{5})}=4.$$
answered Dec 4 at 10:31
farruhota
18.6k2736
answered Dec 4 at 10:31
farruhota
18.6k2736
answered Dec 4 at 10:31
farruhota
18.6k2736
answered Dec 4 at 10:31
farruhota
18.6k2736
18.6k2736
I don't know how should I give you thanks...your answer helps me very much.... ❤❤❤
– RA FI
Dec 4 at 10:57
You are welcome. Note that achille hui hinted to the Pythagoras theorem, implying the triange was right. Good luck.
– farruhota
Dec 4 at 10:59
How could you find out the graph?any mobile apps?
– RA FI
Dec 4 at 11:07
I used desmos for raw graph and edited it.
– farruhota
Dec 4 at 11:09
You can also use graphing calculator.By the way how do u edit this? can I do this in mobile?
– RA FI
Dec 4 at 14:33
|
show 1 more comment
I don't know how should I give you thanks...your answer helps me very much.... ❤❤❤
– RA FI
Dec 4 at 10:57
You are welcome. Note that achille hui hinted to the Pythagoras theorem, implying the triange was right. Good luck.
– farruhota
Dec 4 at 10:59
How could you find out the graph?any mobile apps?
– RA FI
Dec 4 at 11:07
I used desmos for raw graph and edited it.
– farruhota
Dec 4 at 11:09
You can also use graphing calculator.By the way how do u edit this? can I do this in mobile?
– RA FI
Dec 4 at 14:33
I don't know how should I give you thanks...your answer helps me very much.... ❤❤❤
– RA FI
Dec 4 at 10:57
I don't know how should I give you thanks...your answer helps me very much.... ❤❤❤
– RA FI
Dec 4 at 10:57
You are welcome. Note that achille hui hinted to the Pythagoras theorem, implying the triange was right. Good luck.
– farruhota
Dec 4 at 10:59
You are welcome. Note that achille hui hinted to the Pythagoras theorem, implying the triange was right. Good luck.
– farruhota
Dec 4 at 10:59
How could you find out the graph?any mobile apps?
– RA FI
Dec 4 at 11:07
How could you find out the graph?any mobile apps?
– RA FI
Dec 4 at 11:07
I used desmos for raw graph and edited it.
– farruhota
Dec 4 at 11:09
I used desmos for raw graph and edited it.
– farruhota
Dec 4 at 11:09
You can also use graphing calculator.By the way how do u edit this? can I do this in mobile?
– RA FI
Dec 4 at 14:33
You can also use graphing calculator.By the way how do u edit this? can I do this in mobile?
– RA FI
Dec 4 at 14:33
|
show 1 more comment
up vote
5
down vote
With the use of the definition of ellipse $$|PF_1|+|PF_2|=2a=6$$ and the given ratio $$|PF_1|:|PF_2|=2:1,$$ we get $$|PF_1|=4, ; |PF_2|=2$$
Since $F_1F_2=2sqrt 5=sqrt{20}=sqrt{4^2+2^2},$ we have a right triangle $triangle PF_1F_2$ with hypotenuse $F_1F_2.$ The area is $$mathcal{A}=frac 12 cdot |PF_1|cdot |PF_2|=4$$
answered Dec 4 at 10:40
user376343
2,7412822
1
Oh you've done with very simple method. thanks 😍❤
– RA FI
Dec 4 at 11:03
The school exercises often consider special (even trivial) cases :) Oh, I am reading comments and answers only now and I see there are traces indicating the same...
– user376343
Dec 4 at 11:09
Are u talking about the emojis?😅😂
– RA FI
Dec 4 at 14:21
My phone keyboard has this emojis.Are you using from computer?
– RA FI
Dec 4 at 14:52
Thanks, I didn't think this could be used within MSE 🦊
– user376343
Dec 4 at 15:13
add a comment |
up vote
5
down vote
With the use of the definition of ellipse $$|PF_1|+|PF_2|=2a=6$$ and the given ratio $$|PF_1|:|PF_2|=2:1,$$ we get $$|PF_1|=4, ; |PF_2|=2$$
Since $F_1F_2=2sqrt 5=sqrt{20}=sqrt{4^2+2^2},$ we have a right triangle $triangle PF_1F_2$ with hypotenuse $F_1F_2.$ The area is $$mathcal{A}=frac 12 cdot |PF_1|cdot |PF_2|=4$$
answered Dec 4 at 10:40
user376343
2,7412822
1
Oh you've done with very simple method. thanks 😍❤
– RA FI
Dec 4 at 11:03
The school exercises often consider special (even trivial) cases :) Oh, I am reading comments and answers only now and I see there are traces indicating the same...
– user376343
Dec 4 at 11:09
Are u talking about the emojis?😅😂
– RA FI
Dec 4 at 14:21
My phone keyboard has this emojis.Are you using from computer?
– RA FI
Dec 4 at 14:52
Thanks, I didn't think this could be used within MSE 🦊
– user376343
Dec 4 at 15:13
add a comment |
up vote
5
down vote
up vote
5
down vote
With the use of the definition of ellipse $$|PF_1|+|PF_2|=2a=6$$ and the given ratio $$|PF_1|:|PF_2|=2:1,$$ we get $$|PF_1|=4, ; |PF_2|=2$$
Since $F_1F_2=2sqrt 5=sqrt{20}=sqrt{4^2+2^2},$ we have a right triangle $triangle PF_1F_2$ with hypotenuse $F_1F_2.$ The area is $$mathcal{A}=frac 12 cdot |PF_1|cdot |PF_2|=4$$
answered Dec 4 at 10:40
user376343
2,7412822
With the use of the definition of ellipse $$|PF_1|+|PF_2|=2a=6$$ and the given ratio $$|PF_1|:|PF_2|=2:1,$$ we get $$|PF_1|=4, ; |PF_2|=2$$
Since $F_1F_2=2sqrt 5=sqrt{20}=sqrt{4^2+2^2},$ we have a right triangle $triangle PF_1F_2$ with hypotenuse $F_1F_2.$ The area is $$mathcal{A}=frac 12 cdot |PF_1|cdot |PF_2|=4$$
answered Dec 4 at 10:40
user376343
2,7412822
answered Dec 4 at 10:40
user376343
2,7412822
answered Dec 4 at 10:40
user376343
2,7412822
answered Dec 4 at 10:40
user376343
2,7412822
2,7412822
1
Oh you've done with very simple method. thanks 😍❤
– RA FI
Dec 4 at 11:03
The school exercises often consider special (even trivial) cases :) Oh, I am reading comments and answers only now and I see there are traces indicating the same...
– user376343
Dec 4 at 11:09
Are u talking about the emojis?😅😂
– RA FI
Dec 4 at 14:21
My phone keyboard has this emojis.Are you using from computer?
– RA FI
Dec 4 at 14:52
Thanks, I didn't think this could be used within MSE 🦊
– user376343
Dec 4 at 15:13
add a comment |
1
Oh you've done with very simple method. thanks 😍❤
– RA FI
Dec 4 at 11:03
The school exercises often consider special (even trivial) cases :) Oh, I am reading comments and answers only now and I see there are traces indicating the same...
– user376343
Dec 4 at 11:09
Are u talking about the emojis?😅😂
– RA FI
Dec 4 at 14:21
My phone keyboard has this emojis.Are you using from computer?
– RA FI
Dec 4 at 14:52
Thanks, I didn't think this could be used within MSE 🦊
– user376343
Dec 4 at 15:13
1
1
Oh you've done with very simple method. thanks 😍❤
– RA FI
Dec 4 at 11:03
Oh you've done with very simple method. thanks 😍❤
– RA FI
Dec 4 at 11:03
The school exercises often consider special (even trivial) cases :) Oh, I am reading comments and answers only now and I see there are traces indicating the same...
– user376343
Dec 4 at 11:09
The school exercises often consider special (even trivial) cases :) Oh, I am reading comments and answers only now and I see there are traces indicating the same...
– user376343
Dec 4 at 11:09
Are u talking about the emojis?😅😂
– RA FI
Dec 4 at 14:21
Are u talking about the emojis?😅😂
– RA FI
Dec 4 at 14:21
My phone keyboard has this emojis.Are you using from computer?
– RA FI
Dec 4 at 14:52
My phone keyboard has this emojis.Are you using from computer?
– RA FI
Dec 4 at 14:52
Thanks, I didn't think this could be used within MSE 🦊
– user376343
Dec 4 at 15:13
Thanks, I didn't think this could be used within MSE 🦊
– user376343
Dec 4 at 15:13
add a comment |
up vote
3
down vote
Hint:
WLOG $P(3cos t,2sin t)$
$$|PF_1|^2=(3cos t-sqrt5)^2+(2sin t-0)^2=?$$
$$|PF_2|^2=(3cos t+sqrt5)^2+(2sin t-0)^2=?$$
$$dfrac{|PF_1|^2}{|PF_2|^2}=2^2$$
Use $sin^2t=1-cos^2t$ to form a Quadratic Equation in $cos t$
answered Dec 4 at 6:00
lab bhattacharjee
222k15155273
add a comment |
up vote
3
down vote
Hint:
WLOG $P(3cos t,2sin t)$
$$|PF_1|^2=(3cos t-sqrt5)^2+(2sin t-0)^2=?$$
$$|PF_2|^2=(3cos t+sqrt5)^2+(2sin t-0)^2=?$$
$$dfrac{|PF_1|^2}{|PF_2|^2}=2^2$$
Use $sin^2t=1-cos^2t$ to form a Quadratic Equation in $cos t$
answered Dec 4 at 6:00
lab bhattacharjee
222k15155273
add a comment |
up vote
3
down vote
up vote
3
down vote
Hint:
WLOG $P(3cos t,2sin t)$
$$|PF_1|^2=(3cos t-sqrt5)^2+(2sin t-0)^2=?$$
$$|PF_2|^2=(3cos t+sqrt5)^2+(2sin t-0)^2=?$$
$$dfrac{|PF_1|^2}{|PF_2|^2}=2^2$$
Use $sin^2t=1-cos^2t$ to form a Quadratic Equation in $cos t$
answered Dec 4 at 6:00
lab bhattacharjee
222k15155273
Hint:
WLOG $P(3cos t,2sin t)$
$$|PF_1|^2=(3cos t-sqrt5)^2+(2sin t-0)^2=?$$
$$|PF_2|^2=(3cos t+sqrt5)^2+(2sin t-0)^2=?$$
$$dfrac{|PF_1|^2}{|PF_2|^2}=2^2$$
Use $sin^2t=1-cos^2t$ to form a Quadratic Equation in $cos t$
answered Dec 4 at 6:00
lab bhattacharjee
222k15155273
answered Dec 4 at 6:00
lab bhattacharjee
222k15155273
answered Dec 4 at 6:00
lab bhattacharjee
222k15155273
answered Dec 4 at 6:00
lab bhattacharjee
222k15155273
222k15155273
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Hint: $|PF_1|^2 + |PF_2|^2 = 4^2 + 2^2 = 20 = |F_1F_2|^2$
– achille hui
Dec 4 at 8:56
1
guys I didn't get the point... I don't understanding anyone's hint😔.By the way this is my first question.. So I'm really happy for you guys commenting in my question ❤
– RA FI
Dec 4 at 9:23