Find tangets to a ellipse not centered at(0,0) that pass through point P
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I must find the two tangents that pass through the point $(2,7)$ for the ellipse $2x^2+y^2+2x-3y-2=0$ all I was able to was getting $frac{dy}{dx}=frac{(-4x-2)}{(2y-3)}$ and therefore equalizing that with the point I'm given $frac{(-4x-2)}{(2y-3)}=frac{7-y}{2-x}$ so... please help me to find these two tangets
calculus linear-algebra geometry analytic-geometry
asked Dec 10 '18 at 6:32
rorod8rorod8
12
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add a comment |
$begingroup$
I must find the two tangents that pass through the point $(2,7)$ for the ellipse $2x^2+y^2+2x-3y-2=0$ all I was able to was getting $frac{dy}{dx}=frac{(-4x-2)}{(2y-3)}$ and therefore equalizing that with the point I'm given $frac{(-4x-2)}{(2y-3)}=frac{7-y}{2-x}$ so... please help me to find these two tangets
calculus linear-algebra geometry analytic-geometry
asked Dec 10 '18 at 6:32
rorod8rorod8
12
$endgroup$
$begingroup$
Do you know how to find tangents to an ellipse in standard position? This ellipse is just a translation of one.
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– amd
Dec 10 '18 at 7:58
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Yes I know how to do it if it's centered , or at least I have done successfully if it's centered at (0,0)
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– rorod8
Dec 10 '18 at 16:49
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OK, so translate the ellipse so that it’s centered, then translate the two lines back to the original position.
$endgroup$
– amd
Dec 10 '18 at 20:40
add a comment |
$begingroup$
I must find the two tangents that pass through the point $(2,7)$ for the ellipse $2x^2+y^2+2x-3y-2=0$ all I was able to was getting $frac{dy}{dx}=frac{(-4x-2)}{(2y-3)}$ and therefore equalizing that with the point I'm given $frac{(-4x-2)}{(2y-3)}=frac{7-y}{2-x}$ so... please help me to find these two tangets
calculus linear-algebra geometry analytic-geometry
asked Dec 10 '18 at 6:32
rorod8rorod8
12
$endgroup$
I must find the two tangents that pass through the point $(2,7)$ for the ellipse $2x^2+y^2+2x-3y-2=0$ all I was able to was getting $frac{dy}{dx}=frac{(-4x-2)}{(2y-3)}$ and therefore equalizing that with the point I'm given $frac{(-4x-2)}{(2y-3)}=frac{7-y}{2-x}$ so... please help me to find these two tangets
calculus linear-algebra geometry analytic-geometry
calculus linear-algebra geometry analytic-geometry
asked Dec 10 '18 at 6:32
rorod8rorod8
12
asked Dec 10 '18 at 6:32
rorod8rorod8
12
asked Dec 10 '18 at 6:32
rorod8rorod8
12
asked Dec 10 '18 at 6:32
rorod8rorod8
12
asked Dec 10 '18 at 6:32
rorod8rorod8
12
12
$begingroup$
Do you know how to find tangents to an ellipse in standard position? This ellipse is just a translation of one.
$endgroup$
– amd
Dec 10 '18 at 7:58
$begingroup$
Yes I know how to do it if it's centered , or at least I have done successfully if it's centered at (0,0)
$endgroup$
– rorod8
Dec 10 '18 at 16:49
$begingroup$
OK, so translate the ellipse so that it’s centered, then translate the two lines back to the original position.
$endgroup$
– amd
Dec 10 '18 at 20:40
add a comment |
$begingroup$
Do you know how to find tangents to an ellipse in standard position? This ellipse is just a translation of one.
$endgroup$
– amd
Dec 10 '18 at 7:58
$begingroup$
Yes I know how to do it if it's centered , or at least I have done successfully if it's centered at (0,0)
$endgroup$
– rorod8
Dec 10 '18 at 16:49
$begingroup$
OK, so translate the ellipse so that it’s centered, then translate the two lines back to the original position.
$endgroup$
– amd
Dec 10 '18 at 20:40
$begingroup$
Do you know how to find tangents to an ellipse in standard position? This ellipse is just a translation of one.
$endgroup$
– amd
Dec 10 '18 at 7:58
$begingroup$
Do you know how to find tangents to an ellipse in standard position? This ellipse is just a translation of one.
$endgroup$
– amd
Dec 10 '18 at 7:58
$begingroup$
Yes I know how to do it if it's centered , or at least I have done successfully if it's centered at (0,0)
$endgroup$
– rorod8
Dec 10 '18 at 16:49
$begingroup$
Yes I know how to do it if it's centered , or at least I have done successfully if it's centered at (0,0)
$endgroup$
– rorod8
Dec 10 '18 at 16:49
$begingroup$
OK, so translate the ellipse so that it’s centered, then translate the two lines back to the original position.
$endgroup$
– amd
Dec 10 '18 at 20:40
$begingroup$
OK, so translate the ellipse so that it’s centered, then translate the two lines back to the original position.
$endgroup$
– amd
Dec 10 '18 at 20:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
The equation of any straight line passing through $(2,7)$ can be set as $$dfrac{y-7}{x-2}=m$$ where $m$ is the gradient.
Replace the value of $y$ in the equation of the ellipse to form a quadratic equation in $y$
whose two roots represent the ordinate of the intersection
For tangency, the two root must be same.
answered Dec 10 '18 at 6:39
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
$begingroup$
See also : math.stackexchange.com/questions/223462/… math.stackexchange.com/questions/2326909/… math.stackexchange.com/questions/2490384/… math.stackexchange.com/questions/214977/… math.stackexchange.com/questions/780603/…
$endgroup$
– lab bhattacharjee
Dec 10 '18 at 6:42
$begingroup$
thx!!!! I had thought of resolving it like that, like equalizing both equations and then set the discriminant to 0, then solve it like a quadratic equation for m, but I had had so much trouble that I abandoned algebra and tried to do it with calculus , but it finally worked out prntscr.com/lt6vxz
$endgroup$
– rorod8
Dec 10 '18 at 17:53
add a comment |
$begingroup$
Since you know how to find the tangents to an ellipse that’s centered on the origin, a way to approach this problem is to translate this ellipse so that its center is at the origin (and of course translate $(2,7)$ by the same amount), find the tangents, and then translate them back.
The ellipse’s center can by found in various ways such as computing partial derivatives and setting them equal to zero: the center of the ellipse is the solution to the equations $$begin{align} 4x+2 &= 0 \ 2y-3 &= 0. end{align}$$ Translate the origin to the center point $(x_c,y_c)$ by making the substitutions $xto x+x_c$ and $yto y+y_c$ in its equation. A short cut: doing this will leave the quadratic terms alone, eliminate the first-degree terms in $x$ and $y$ and replace the constant term with the value of the left-hand side of the original equation at $(x_c,y_c)$, i.e., the translated equation will be $$2x^2+y^2+(2x_c^2+y_c^2+2x_c-3y_c-2)=0.$$ Find the tangents to this ellipse that go through the point $(2-x_c,y-y_c)$. Once you’ve done this, translate the equations of the lines back to the original coordinate system with the substitutions $xto x-x_c$ and $yto y-y_c$.
A more direct approach is to use the polar of the point $(2,7)$, which is a line that includes the chord of contact of the tangents through the point. The equation of the polar of a point $(x_0,y_0)$ can be found mechanically by making the following substitutions in the equation of the conic: $$x^2 to xx_0 \ y^2 to yy_0 \ xy to frac12(xy_0+x_0y) \ x to frac12(x+x_0) \ y to frac12(y+y_0).$$ Find the intersections of the resulting line with the ellipse and derive the equations of the lines through these points and $(2,7)$.
answered Dec 13 '18 at 1:16
amdamd
31.1k21051
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint:
The equation of any straight line passing through $(2,7)$ can be set as $$dfrac{y-7}{x-2}=m$$ where $m$ is the gradient.
Replace the value of $y$ in the equation of the ellipse to form a quadratic equation in $y$
whose two roots represent the ordinate of the intersection
For tangency, the two root must be same.
answered Dec 10 '18 at 6:39
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
$begingroup$
See also : math.stackexchange.com/questions/223462/… math.stackexchange.com/questions/2326909/… math.stackexchange.com/questions/2490384/… math.stackexchange.com/questions/214977/… math.stackexchange.com/questions/780603/…
$endgroup$
– lab bhattacharjee
Dec 10 '18 at 6:42
$begingroup$
thx!!!! I had thought of resolving it like that, like equalizing both equations and then set the discriminant to 0, then solve it like a quadratic equation for m, but I had had so much trouble that I abandoned algebra and tried to do it with calculus , but it finally worked out prntscr.com/lt6vxz
$endgroup$
– rorod8
Dec 10 '18 at 17:53
add a comment |
$begingroup$
Hint:
The equation of any straight line passing through $(2,7)$ can be set as $$dfrac{y-7}{x-2}=m$$ where $m$ is the gradient.
Replace the value of $y$ in the equation of the ellipse to form a quadratic equation in $y$
whose two roots represent the ordinate of the intersection
For tangency, the two root must be same.
answered Dec 10 '18 at 6:39
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
$begingroup$
See also : math.stackexchange.com/questions/223462/… math.stackexchange.com/questions/2326909/… math.stackexchange.com/questions/2490384/… math.stackexchange.com/questions/214977/… math.stackexchange.com/questions/780603/…
$endgroup$
– lab bhattacharjee
Dec 10 '18 at 6:42
$begingroup$
thx!!!! I had thought of resolving it like that, like equalizing both equations and then set the discriminant to 0, then solve it like a quadratic equation for m, but I had had so much trouble that I abandoned algebra and tried to do it with calculus , but it finally worked out prntscr.com/lt6vxz
$endgroup$
– rorod8
Dec 10 '18 at 17:53
add a comment |
$begingroup$
Hint:
The equation of any straight line passing through $(2,7)$ can be set as $$dfrac{y-7}{x-2}=m$$ where $m$ is the gradient.
Replace the value of $y$ in the equation of the ellipse to form a quadratic equation in $y$
whose two roots represent the ordinate of the intersection
For tangency, the two root must be same.
answered Dec 10 '18 at 6:39
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
Hint:
The equation of any straight line passing through $(2,7)$ can be set as $$dfrac{y-7}{x-2}=m$$ where $m$ is the gradient.
Replace the value of $y$ in the equation of the ellipse to form a quadratic equation in $y$
whose two roots represent the ordinate of the intersection
For tangency, the two root must be same.
answered Dec 10 '18 at 6:39
lab bhattacharjeelab bhattacharjee
227k15158276
answered Dec 10 '18 at 6:39
lab bhattacharjeelab bhattacharjee
227k15158276
answered Dec 10 '18 at 6:39
lab bhattacharjeelab bhattacharjee
227k15158276
answered Dec 10 '18 at 6:39
lab bhattacharjeelab bhattacharjee
227k15158276
227k15158276
$begingroup$
See also : math.stackexchange.com/questions/223462/… math.stackexchange.com/questions/2326909/… math.stackexchange.com/questions/2490384/… math.stackexchange.com/questions/214977/… math.stackexchange.com/questions/780603/…
$endgroup$
– lab bhattacharjee
Dec 10 '18 at 6:42
$begingroup$
thx!!!! I had thought of resolving it like that, like equalizing both equations and then set the discriminant to 0, then solve it like a quadratic equation for m, but I had had so much trouble that I abandoned algebra and tried to do it with calculus , but it finally worked out prntscr.com/lt6vxz
$endgroup$
– rorod8
Dec 10 '18 at 17:53
add a comment |
$begingroup$
See also : math.stackexchange.com/questions/223462/… math.stackexchange.com/questions/2326909/… math.stackexchange.com/questions/2490384/… math.stackexchange.com/questions/214977/… math.stackexchange.com/questions/780603/…
$endgroup$
– lab bhattacharjee
Dec 10 '18 at 6:42
$begingroup$
thx!!!! I had thought of resolving it like that, like equalizing both equations and then set the discriminant to 0, then solve it like a quadratic equation for m, but I had had so much trouble that I abandoned algebra and tried to do it with calculus , but it finally worked out prntscr.com/lt6vxz
$endgroup$
– rorod8
Dec 10 '18 at 17:53
$begingroup$
See also : math.stackexchange.com/questions/223462/… math.stackexchange.com/questions/2326909/… math.stackexchange.com/questions/2490384/… math.stackexchange.com/questions/214977/… math.stackexchange.com/questions/780603/…
$endgroup$
– lab bhattacharjee
Dec 10 '18 at 6:42
$begingroup$
See also : math.stackexchange.com/questions/223462/… math.stackexchange.com/questions/2326909/… math.stackexchange.com/questions/2490384/… math.stackexchange.com/questions/214977/… math.stackexchange.com/questions/780603/…
$endgroup$
– lab bhattacharjee
Dec 10 '18 at 6:42
$begingroup$
thx!!!! I had thought of resolving it like that, like equalizing both equations and then set the discriminant to 0, then solve it like a quadratic equation for m, but I had had so much trouble that I abandoned algebra and tried to do it with calculus , but it finally worked out prntscr.com/lt6vxz
$endgroup$
– rorod8
Dec 10 '18 at 17:53
$begingroup$
thx!!!! I had thought of resolving it like that, like equalizing both equations and then set the discriminant to 0, then solve it like a quadratic equation for m, but I had had so much trouble that I abandoned algebra and tried to do it with calculus , but it finally worked out prntscr.com/lt6vxz
$endgroup$
– rorod8
Dec 10 '18 at 17:53
add a comment |
$begingroup$
Since you know how to find the tangents to an ellipse that’s centered on the origin, a way to approach this problem is to translate this ellipse so that its center is at the origin (and of course translate $(2,7)$ by the same amount), find the tangents, and then translate them back.
The ellipse’s center can by found in various ways such as computing partial derivatives and setting them equal to zero: the center of the ellipse is the solution to the equations $$begin{align} 4x+2 &= 0 \ 2y-3 &= 0. end{align}$$ Translate the origin to the center point $(x_c,y_c)$ by making the substitutions $xto x+x_c$ and $yto y+y_c$ in its equation. A short cut: doing this will leave the quadratic terms alone, eliminate the first-degree terms in $x$ and $y$ and replace the constant term with the value of the left-hand side of the original equation at $(x_c,y_c)$, i.e., the translated equation will be $$2x^2+y^2+(2x_c^2+y_c^2+2x_c-3y_c-2)=0.$$ Find the tangents to this ellipse that go through the point $(2-x_c,y-y_c)$. Once you’ve done this, translate the equations of the lines back to the original coordinate system with the substitutions $xto x-x_c$ and $yto y-y_c$.
A more direct approach is to use the polar of the point $(2,7)$, which is a line that includes the chord of contact of the tangents through the point. The equation of the polar of a point $(x_0,y_0)$ can be found mechanically by making the following substitutions in the equation of the conic: $$x^2 to xx_0 \ y^2 to yy_0 \ xy to frac12(xy_0+x_0y) \ x to frac12(x+x_0) \ y to frac12(y+y_0).$$ Find the intersections of the resulting line with the ellipse and derive the equations of the lines through these points and $(2,7)$.
answered Dec 13 '18 at 1:16
amdamd
31.1k21051
$endgroup$
add a comment |
$begingroup$
Since you know how to find the tangents to an ellipse that’s centered on the origin, a way to approach this problem is to translate this ellipse so that its center is at the origin (and of course translate $(2,7)$ by the same amount), find the tangents, and then translate them back.
The ellipse’s center can by found in various ways such as computing partial derivatives and setting them equal to zero: the center of the ellipse is the solution to the equations $$begin{align} 4x+2 &= 0 \ 2y-3 &= 0. end{align}$$ Translate the origin to the center point $(x_c,y_c)$ by making the substitutions $xto x+x_c$ and $yto y+y_c$ in its equation. A short cut: doing this will leave the quadratic terms alone, eliminate the first-degree terms in $x$ and $y$ and replace the constant term with the value of the left-hand side of the original equation at $(x_c,y_c)$, i.e., the translated equation will be $$2x^2+y^2+(2x_c^2+y_c^2+2x_c-3y_c-2)=0.$$ Find the tangents to this ellipse that go through the point $(2-x_c,y-y_c)$. Once you’ve done this, translate the equations of the lines back to the original coordinate system with the substitutions $xto x-x_c$ and $yto y-y_c$.
A more direct approach is to use the polar of the point $(2,7)$, which is a line that includes the chord of contact of the tangents through the point. The equation of the polar of a point $(x_0,y_0)$ can be found mechanically by making the following substitutions in the equation of the conic: $$x^2 to xx_0 \ y^2 to yy_0 \ xy to frac12(xy_0+x_0y) \ x to frac12(x+x_0) \ y to frac12(y+y_0).$$ Find the intersections of the resulting line with the ellipse and derive the equations of the lines through these points and $(2,7)$.
answered Dec 13 '18 at 1:16
amdamd
31.1k21051
$endgroup$
add a comment |
$begingroup$
Since you know how to find the tangents to an ellipse that’s centered on the origin, a way to approach this problem is to translate this ellipse so that its center is at the origin (and of course translate $(2,7)$ by the same amount), find the tangents, and then translate them back.
The ellipse’s center can by found in various ways such as computing partial derivatives and setting them equal to zero: the center of the ellipse is the solution to the equations $$begin{align} 4x+2 &= 0 \ 2y-3 &= 0. end{align}$$ Translate the origin to the center point $(x_c,y_c)$ by making the substitutions $xto x+x_c$ and $yto y+y_c$ in its equation. A short cut: doing this will leave the quadratic terms alone, eliminate the first-degree terms in $x$ and $y$ and replace the constant term with the value of the left-hand side of the original equation at $(x_c,y_c)$, i.e., the translated equation will be $$2x^2+y^2+(2x_c^2+y_c^2+2x_c-3y_c-2)=0.$$ Find the tangents to this ellipse that go through the point $(2-x_c,y-y_c)$. Once you’ve done this, translate the equations of the lines back to the original coordinate system with the substitutions $xto x-x_c$ and $yto y-y_c$.
A more direct approach is to use the polar of the point $(2,7)$, which is a line that includes the chord of contact of the tangents through the point. The equation of the polar of a point $(x_0,y_0)$ can be found mechanically by making the following substitutions in the equation of the conic: $$x^2 to xx_0 \ y^2 to yy_0 \ xy to frac12(xy_0+x_0y) \ x to frac12(x+x_0) \ y to frac12(y+y_0).$$ Find the intersections of the resulting line with the ellipse and derive the equations of the lines through these points and $(2,7)$.
answered Dec 13 '18 at 1:16
amdamd
31.1k21051
$endgroup$
Since you know how to find the tangents to an ellipse that’s centered on the origin, a way to approach this problem is to translate this ellipse so that its center is at the origin (and of course translate $(2,7)$ by the same amount), find the tangents, and then translate them back.
The ellipse’s center can by found in various ways such as computing partial derivatives and setting them equal to zero: the center of the ellipse is the solution to the equations $$begin{align} 4x+2 &= 0 \ 2y-3 &= 0. end{align}$$ Translate the origin to the center point $(x_c,y_c)$ by making the substitutions $xto x+x_c$ and $yto y+y_c$ in its equation. A short cut: doing this will leave the quadratic terms alone, eliminate the first-degree terms in $x$ and $y$ and replace the constant term with the value of the left-hand side of the original equation at $(x_c,y_c)$, i.e., the translated equation will be $$2x^2+y^2+(2x_c^2+y_c^2+2x_c-3y_c-2)=0.$$ Find the tangents to this ellipse that go through the point $(2-x_c,y-y_c)$. Once you’ve done this, translate the equations of the lines back to the original coordinate system with the substitutions $xto x-x_c$ and $yto y-y_c$.
A more direct approach is to use the polar of the point $(2,7)$, which is a line that includes the chord of contact of the tangents through the point. The equation of the polar of a point $(x_0,y_0)$ can be found mechanically by making the following substitutions in the equation of the conic: $$x^2 to xx_0 \ y^2 to yy_0 \ xy to frac12(xy_0+x_0y) \ x to frac12(x+x_0) \ y to frac12(y+y_0).$$ Find the intersections of the resulting line with the ellipse and derive the equations of the lines through these points and $(2,7)$.
answered Dec 13 '18 at 1:16
amdamd
31.1k21051
edited Dec 13 '18 at 1:34
answered Dec 13 '18 at 1:16
amdamd
31.1k21051
answered Dec 13 '18 at 1:16
amdamd
31.1k21051
answered Dec 13 '18 at 1:16
amdamd
31.1k21051
31.1k21051
add a comment |
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Do you know how to find tangents to an ellipse in standard position? This ellipse is just a translation of one.
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– amd
Dec 10 '18 at 7:58
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Yes I know how to do it if it's centered , or at least I have done successfully if it's centered at (0,0)
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– rorod8
Dec 10 '18 at 16:49
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OK, so translate the ellipse so that it’s centered, then translate the two lines back to the original position.
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– amd
Dec 10 '18 at 20:40