Mixing
Express the coordinates $(x_1,y_1)$ in terms of $(x_0,y_0)$ and $t$
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Question: Let $(x_0,y_0)$ be a point of the curve $y^2=ax^2+bx+c$ and $t$ is the slope of the line passing through $(x_0,y_0)$ and intersecting this curve in the point $(x_1,y_1)$. Express the coordinates $(x_1,y_1)$ in terms of $(x_0,y_0)$ and $t$.
My attempt: Suppose the equation of line be
$$y-y_0=t(x-x_0)$$
Since the line intersect the curve at the points $(x_0,y_0)$ and $(x_1,y_1)$, so
$$[t(x-x_0)+y_0]^2=ax^2+bx+c$$
After simplification, I have obtained
$$(a-t^2)x^2+(b+2x_0t^2-2ty_0)x+(c-x_0^2t^2+2x_0y_0t-y_0^2)=0$$
How do I simplify this in order to find the coordinate $(x_1,y_1)$?
analytic-geometry
asked Jan 2 at 15:37
weilam06weilam06
16011
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add a comment |
$begingroup$
Question: Let $(x_0,y_0)$ be a point of the curve $y^2=ax^2+bx+c$ and $t$ is the slope of the line passing through $(x_0,y_0)$ and intersecting this curve in the point $(x_1,y_1)$. Express the coordinates $(x_1,y_1)$ in terms of $(x_0,y_0)$ and $t$.
My attempt: Suppose the equation of line be
$$y-y_0=t(x-x_0)$$
Since the line intersect the curve at the points $(x_0,y_0)$ and $(x_1,y_1)$, so
$$[t(x-x_0)+y_0]^2=ax^2+bx+c$$
After simplification, I have obtained
$$(a-t^2)x^2+(b+2x_0t^2-2ty_0)x+(c-x_0^2t^2+2x_0y_0t-y_0^2)=0$$
How do I simplify this in order to find the coordinate $(x_1,y_1)$?
analytic-geometry
asked Jan 2 at 15:37
weilam06weilam06
16011
$endgroup$
add a comment |
$begingroup$
Question: Let $(x_0,y_0)$ be a point of the curve $y^2=ax^2+bx+c$ and $t$ is the slope of the line passing through $(x_0,y_0)$ and intersecting this curve in the point $(x_1,y_1)$. Express the coordinates $(x_1,y_1)$ in terms of $(x_0,y_0)$ and $t$.
My attempt: Suppose the equation of line be
$$y-y_0=t(x-x_0)$$
Since the line intersect the curve at the points $(x_0,y_0)$ and $(x_1,y_1)$, so
$$[t(x-x_0)+y_0]^2=ax^2+bx+c$$
After simplification, I have obtained
$$(a-t^2)x^2+(b+2x_0t^2-2ty_0)x+(c-x_0^2t^2+2x_0y_0t-y_0^2)=0$$
How do I simplify this in order to find the coordinate $(x_1,y_1)$?
analytic-geometry
asked Jan 2 at 15:37
weilam06weilam06
16011
$endgroup$
Question: Let $(x_0,y_0)$ be a point of the curve $y^2=ax^2+bx+c$ and $t$ is the slope of the line passing through $(x_0,y_0)$ and intersecting this curve in the point $(x_1,y_1)$. Express the coordinates $(x_1,y_1)$ in terms of $(x_0,y_0)$ and $t$.
My attempt: Suppose the equation of line be
$$y-y_0=t(x-x_0)$$
Since the line intersect the curve at the points $(x_0,y_0)$ and $(x_1,y_1)$, so
$$[t(x-x_0)+y_0]^2=ax^2+bx+c$$
After simplification, I have obtained
$$(a-t^2)x^2+(b+2x_0t^2-2ty_0)x+(c-x_0^2t^2+2x_0y_0t-y_0^2)=0$$
How do I simplify this in order to find the coordinate $(x_1,y_1)$?
analytic-geometry
analytic-geometry
asked Jan 2 at 15:37
weilam06weilam06
16011
asked Jan 2 at 15:37
weilam06weilam06
16011
asked Jan 2 at 15:37
weilam06weilam06
16011
asked Jan 2 at 15:37
weilam06weilam06
16011
asked Jan 2 at 15:37
weilam06weilam06
16011
16011
add a comment |
add a comment |
1 Answer
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You know $(x_0,y_0)$ intersects the curve. So $$y_0^2=ax_0^2+bx_0+c$$You have also found $$[t(x-x_0)+y_0]^2=ax^2+bx+c$$Which is also solved by $(x_1,y_1)$, so $$[t(x_1-x_0)+y_0]^2=ax_1^2+bx_1+c\implies [t(x_1-x_0)+y_0]^2-y_0^2=a(x_1^2-x_0^2)+b(x_1-x_0)\t(x_1-x_0)(t(x_1-x_0)+2y_0)=a(x_1-x_0)(x_1+x_0)+b(x_1-x_0)$$
Let $z=x_1-x_0$. $$tz(tz+2y_0)=az(z+2x_0)+bz\(t^2-a)z^2+(2y_0t-2x_0a-b)z=0$$Since $x_1ne x_0$, we have $zne0$.
$$z=x_1-x_0=frac{2y_0t-2x_0a-b}{a-t^2}$$
You now have $x_1$ in terms of $(x_0,y_0)$. You can then use $y_1^2=ax_1^2+bx_1+c$ to find $y_1$ in terms of these quantities too.
answered Jan 2 at 15:51
John DoeJohn Doe
11.1k11238
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1 Answer
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1 Answer
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$begingroup$
You know $(x_0,y_0)$ intersects the curve. So $$y_0^2=ax_0^2+bx_0+c$$You have also found $$[t(x-x_0)+y_0]^2=ax^2+bx+c$$Which is also solved by $(x_1,y_1)$, so $$[t(x_1-x_0)+y_0]^2=ax_1^2+bx_1+c\implies [t(x_1-x_0)+y_0]^2-y_0^2=a(x_1^2-x_0^2)+b(x_1-x_0)\t(x_1-x_0)(t(x_1-x_0)+2y_0)=a(x_1-x_0)(x_1+x_0)+b(x_1-x_0)$$
Let $z=x_1-x_0$. $$tz(tz+2y_0)=az(z+2x_0)+bz\(t^2-a)z^2+(2y_0t-2x_0a-b)z=0$$Since $x_1ne x_0$, we have $zne0$.
$$z=x_1-x_0=frac{2y_0t-2x_0a-b}{a-t^2}$$
You now have $x_1$ in terms of $(x_0,y_0)$. You can then use $y_1^2=ax_1^2+bx_1+c$ to find $y_1$ in terms of these quantities too.
answered Jan 2 at 15:51
John DoeJohn Doe
11.1k11238
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add a comment |
$begingroup$
You know $(x_0,y_0)$ intersects the curve. So $$y_0^2=ax_0^2+bx_0+c$$You have also found $$[t(x-x_0)+y_0]^2=ax^2+bx+c$$Which is also solved by $(x_1,y_1)$, so $$[t(x_1-x_0)+y_0]^2=ax_1^2+bx_1+c\implies [t(x_1-x_0)+y_0]^2-y_0^2=a(x_1^2-x_0^2)+b(x_1-x_0)\t(x_1-x_0)(t(x_1-x_0)+2y_0)=a(x_1-x_0)(x_1+x_0)+b(x_1-x_0)$$
Let $z=x_1-x_0$. $$tz(tz+2y_0)=az(z+2x_0)+bz\(t^2-a)z^2+(2y_0t-2x_0a-b)z=0$$Since $x_1ne x_0$, we have $zne0$.
$$z=x_1-x_0=frac{2y_0t-2x_0a-b}{a-t^2}$$
You now have $x_1$ in terms of $(x_0,y_0)$. You can then use $y_1^2=ax_1^2+bx_1+c$ to find $y_1$ in terms of these quantities too.
answered Jan 2 at 15:51
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
You know $(x_0,y_0)$ intersects the curve. So $$y_0^2=ax_0^2+bx_0+c$$You have also found $$[t(x-x_0)+y_0]^2=ax^2+bx+c$$Which is also solved by $(x_1,y_1)$, so $$[t(x_1-x_0)+y_0]^2=ax_1^2+bx_1+c\implies [t(x_1-x_0)+y_0]^2-y_0^2=a(x_1^2-x_0^2)+b(x_1-x_0)\t(x_1-x_0)(t(x_1-x_0)+2y_0)=a(x_1-x_0)(x_1+x_0)+b(x_1-x_0)$$
Let $z=x_1-x_0$. $$tz(tz+2y_0)=az(z+2x_0)+bz\(t^2-a)z^2+(2y_0t-2x_0a-b)z=0$$Since $x_1ne x_0$, we have $zne0$.
$$z=x_1-x_0=frac{2y_0t-2x_0a-b}{a-t^2}$$
You now have $x_1$ in terms of $(x_0,y_0)$. You can then use $y_1^2=ax_1^2+bx_1+c$ to find $y_1$ in terms of these quantities too.
answered Jan 2 at 15:51
John DoeJohn Doe
11.1k11238
$endgroup$
You know $(x_0,y_0)$ intersects the curve. So $$y_0^2=ax_0^2+bx_0+c$$You have also found $$[t(x-x_0)+y_0]^2=ax^2+bx+c$$Which is also solved by $(x_1,y_1)$, so $$[t(x_1-x_0)+y_0]^2=ax_1^2+bx_1+c\implies [t(x_1-x_0)+y_0]^2-y_0^2=a(x_1^2-x_0^2)+b(x_1-x_0)\t(x_1-x_0)(t(x_1-x_0)+2y_0)=a(x_1-x_0)(x_1+x_0)+b(x_1-x_0)$$
Let $z=x_1-x_0$. $$tz(tz+2y_0)=az(z+2x_0)+bz\(t^2-a)z^2+(2y_0t-2x_0a-b)z=0$$Since $x_1ne x_0$, we have $zne0$.
$$z=x_1-x_0=frac{2y_0t-2x_0a-b}{a-t^2}$$
You now have $x_1$ in terms of $(x_0,y_0)$. You can then use $y_1^2=ax_1^2+bx_1+c$ to find $y_1$ in terms of these quantities too.
answered Jan 2 at 15:51
John DoeJohn Doe
11.1k11238
edited Jan 2 at 16:07
answered Jan 2 at 15:51
John DoeJohn Doe
11.1k11238
answered Jan 2 at 15:51
John DoeJohn Doe
11.1k11238
answered Jan 2 at 15:51
John DoeJohn Doe
11.1k11238
11.1k11238
add a comment |
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