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Question Number 117739 by bemath last updated on 13/Oct/20

$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{circle}$$$$\mathrm{with}\mathrm{radius}4\sqrt{2}\mathrm{that}\mathrm{can}\mathrm{be}$$$$\mathrm{inscribed}\mathrm{in}\mathrm{the}\mathrm{parabola}$$$$\mathrm{y}={\mathrm{x}}^2-16\mathrm{x}+128?$$

Answered by bobhans last updated on 13/Oct/20

$$\mathrm{for}\mathrm{symetry}\mathrm{reasons},\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{circle}$$$$\mathrm{will}\mathrm{lie}\mathrm{on}\mathrm{the}\mathrm{axis}\mathrm{of}\mathrm{the}\mathrm{parabola},\mathrm{say}$$$$\mathrm{it}\mathrm{center}\mathrm{is}(8,\mathrm{u})\mathrm{and}\mathrm{the}\mathrm{equation}\mathrm{is}$$$${(\mathrm{x}-8)}^2+{(\mathrm{y}-\mathrm{u})}^2={\left(4\sqrt{2}\right)}^2$$$${(\mathrm{x}-8)}^2+{(\mathrm{y}-\mathrm{u})}^2=32$$$${\mathrm{x}}^2-16\mathrm{x}=-{(\mathrm{y}-\mathrm{u})}^2-32$$$$\mathrm{if}\mathrm{we}\mathrm{subtitute}\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}$$$$\mathrm{parabola}\mathrm{gives}\mathrm{y}=-{(\mathrm{y}-\mathrm{u})}^2-32$$$$\mathrm{or}{\mathrm{y}}^2+(1-2\mathrm{u})\mathrm{y}+{\mathrm{u}}^2-96=0$$$$\mathrm{this}\mathrm{should}\mathrm{be}\mathrm{have}\mathrm{one}\mathrm{roots},\mathrm{so}\mathrm{we}$$$$\mathrm{get}{(1-2\mathrm{u})}^2-4({\mathrm{u}}^2-96)=0$$$$1-4\mathrm{u}+{4\mathrm{u}}^2-{4\mathrm{u}}^2+384=0$$$$\iff \mathrm{u}=\frac{385}{4}.\mathrm{Thus}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{the}\mathrm{circle}$$$$\mathrm{is}(8,\frac{385}{4})$$$$$$

Commented by bemath last updated on 13/Oct/20

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