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Question Number 199522 by cherokeesay last updated on 04/Nov/23

Answered by cortano12 last updated on 05/Nov/23

$$\left({\mathrm{g}}_1\right)\equiv \mathrm{y}=\mathrm{ax}+\mathrm{b}\left(\mathrm{tangent}\mathrm{line}\right)$$$$\mathrm{where}\mathrm{a}=\frac{1}{4\left(\frac{1}{4}\right)}=1$$$$\mathrm{and}\mathrm{b}=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$$$$\therefore \mathrm{y}=\mathrm{x}+\frac{1}{8}$$$$\left({\mathrm{h}}_1\right)\equiv \mathrm{y}=\mathrm{px}+\mathrm{q}\left(\mathrm{normal}\mathrm{line}\right)$$$$\mathrm{where}\mathrm{p}=-\frac{1}{\mathrm{a}}=-\frac{1}{1}=-1$$$$\mathrm{and}\mathrm{q}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$$$$\therefore \mathrm{y}=-\mathrm{x}+\frac{3}{8}$$$$(\ast )\mathrm{normal}\mathrm{cuts}\mathrm{the}\mathrm{parabola}$$$$\Rightarrow {(-\mathrm{x}+\frac{3}{8})}^2=\frac{1}{2}\mathrm{x}$$$${\mathrm{x}}^2-\frac{3}{4}\mathrm{x}+\frac{9}{64}=\frac{1}{2}\mathrm{x}$$$${\mathrm{x}}^2-\frac{5}{4}\mathrm{x}+\frac{9}{64}=0$$$$(\mathrm{x}-\frac{1}{8})(\mathrm{x}-\frac{9}{8})=0$$$$\mathrm{the}\mathrm{other}\mathrm{point}\mathrm{is}\mathrm{R}(\frac{9}{8},-\frac{3}{4})$$$$\mathrm{equation}\mathrm{of}\mathrm{tangent}\mathrm{at}\mathrm{R}(\frac{9}{8},-\frac{3}{4})$$$$\mathrm{is}\mathrm{y}=\mathrm{rx}+\mathrm{s},\mathrm{where}\mathrm{r}=\frac{1}{4(-\frac{3}{4})}=-\frac{1}{3}$$$$\mathrm{and}\mathrm{s}=-\frac{3}{4}+\frac{1}{3}.\frac{9}{8}=-\frac{3}{4}+\frac{3}{8}=-\frac{3}{8}$$$$\therefore \mathrm{y}=-\frac{1}{3}\mathrm{x}-\frac{3}{8}$$$$\mathrm{Let}\alpha \mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{two}$$$$\mathrm{tangent}\Rightarrow \tan \alpha =\mid \frac{1-(-\frac{1}{3})}{1+1.(-\frac{1}{3})}\mid$$$$\tan \alpha =2\Rightarrow \alpha =\arctan \left(2\right)$$

Commented by cherokeesay last updated on 05/Nov/23

$$thankyousir!$$

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