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Question Number 93855 by Ar Brandon last updated on 15/May/20

$$\mathrm{At}\mathrm{what}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{parabola}\mathrm{y}={\mathrm{x}}^2$$$$\mathrm{will}\mathrm{the}\mathrm{tangent}$$$$\mathrm{a}\mathrm{\setminus }\mathrm{be}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{y}=4\mathrm{x}-5$$$$\mathrm{b}\mathrm{\setminus }\mathrm{be}\mathrm{perpendicular}\mathrm{to}2\mathrm{x}-6\mathrm{y}+5=0$$$$\mathrm{c}\mathrm{\setminus }\mathrm{make}\mathrm{an}\mathrm{angle}\mathrm{of}{45}^{°}\mathrm{with}3\mathrm{x}-\mathrm{y}+1=0$$

Answered by Kunal12588 last updated on 15/May/20

$$a⟩when2x=4\Rightarrow x=2;point(2,4)$$$$b⟩when\frac{-1}{2x}=\frac{1}{3}\Rightarrow x=\frac{-3}{2};point(-\frac{3}{2},\frac{9}{4})$$$$c⟩when1=\frac{2x-3}{1+6x}\Rightarrow 4x=-4\Rightarrow x=-1;point(-1,1)$$

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