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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 $(x_1,y_1)$?

weilam06
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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 $x_1\ne x_0$, we have $z\ne0$. $$z=x_1-x_0=\frac{2y_0t-2x_0a-b}{a-t^2}$$

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.

John Doe
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