For the parabola y^2=4ax, prove that the locus of point of intersection of two tangents which intercept a given distance 4C on the tangent at the vertex is again a parabola.?

Dear Student,

$$\begin{gathered} The \tan gent at the vertex is y axis (equation x = 0) \\ so its intersection points with each of the other 2 \tan gents \\ will have abscissa 0 \\ and according that two ordinates \\ are \\ \frac{2a{x}_1}{y_1}=\frac{y_1}{2} and \frac{2a{x}_2}{y_2}=\frac{y_2}{2} \\ because y^2=4ax \\ The dis\tan ce between them is \left|\frac{y_1}{2}-\frac{y_2}{2}\right|=4C \\ \left|y_1-y_2\right|=8C \\ sqauring both side \\ (y_1-y_2{)}^2=(8C{)}^2 \\ {\left(y_1+y_2\right)}^2-4y_1{y}_2=64C^2 \\ 4y^2-16ax=64C^2 \left[taking y_1+y_2=2y and y_1{y}_2=y^2 and y^2=4ax\right] \\ {y}^2-4ax=16C^2 \\ {y}^2=16C^2+4ax \\ {y}^2=16C^2+4ax \\ {\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{4}\mathit{a}\left(\mathbf{x}\mathbf{+}\frac{\mathbf{4}{\mathbf{C}}^{\mathbf{2}}}{\mathbf{a}}\right)\mathbf{ }\mathit{E}\mathit{q}\mathit{u}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{P}\mathit{a}\mathit{r}\mathit{a}\mathit{b}\mathit{o}\mathit{l}\mathit{a} \end{gathered}$$


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