Question

The equation of the line passing through the points (at1^2 , 2at1), (at2^2 , 2at2) is​

Answer 1

$\textbf{Given:}$

$\text{Points are}\;(a\,{t_1}^2,2a\,t_1)\;\text{and}\;(a\,{t_2}^2,2a\,t_2)$

$\textbf{To find:}$

$\text{The equation of line joining the given two points}$

$\textbf{Solution:}$

$\text{We know that,}$

$\text{The equation of line joining (x_1,y_1) and (x_2,y_2) is}$

$\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}$

$\dfrac{y-2a\,t_1}{2a\,t_2-2a\,t_1}=\dfrac{x-a\,{t_1}^2}{a\,{t_2}^2-a\,{t_1}^2}$

$\dfrac{y-2a\,t_1}{2a(t_2-\,t_1)}=\dfrac{x-a\,{t_1}^2}{a({t_2}^2-{t_1}^2)}$

$\dfrac{y-2a\,t_1}{2(t_2-\,t_1)}=\dfrac{x-a\,{t_1}^2}{({t_2}-{t_1})({t_2}+{t_1})}$

$\dfrac{y-2a\,t_1}{2}=\dfrac{x-a\,{t_1}^2}{({t_2}+{t_1})}$

$(y-2a\,t_1)({t_2}+{t_1})=2(x-a\,{t_1}^2)$

$({t_1}+{t_2})y-2a\,t_1({t_1}+{t_2})=2x-2a\,{t_1}^2$

$2a\,{t_1}^2-2a\,t_1({t_1}+{t_2})=2x-({t_1}+{t_2})y$

$2a\,{t_1}^2-2a\,{t_1}^2-2a\,{t_1}\,{t_2}=2x-({t_1}+{t_2})y$

$-2a\,{t_1}\,{t_2}=2x-({t_1}+{t_2})y$

$\text{Rearranging terms we get}$

$\bf\,2x-({t_1}+{t_2})y+2a\,{t_1}\,{t_2}=0$

$\textbf{Answer:}$

$\textbf{The required line is}$

$\boxed{\bf\,2x-({t_1}+{t_2})y+2a\,{t_1}\,{t_2}=0}$

Answer 2

Answer:

above is ur answer dear friends