Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 130213 by benjo_mathlover last updated on 23/Jan/21

$$\mathrm{The}\mathrm{closest}\mathrm{distance}\mathrm{from}\mathrm{the}\mathrm{point}\mathrm{on}$$$$\mathrm{the}\mathrm{ellipse}{2\mathrm{x}}^2-{\mathrm{y}}^2=8\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{y}=5\mathrm{x}$$$$\mathrm{is}\mathrm{\_}\mathrm{\_}$$

Answered by liberty last updated on 23/Jan/21

$$\mathrm{gradient}\mathrm{tangent}\mathrm{line}\mathrm{of}$$$$\mathrm{ellipse}:4\mathrm{x}-{2\mathrm{yy}}^{\prime }=0$$$$\Rightarrow {\mathrm{y}}^{\prime }=\frac{2\mathrm{x}}{\mathrm{y}}\mathrm{equals}\mathrm{to}5;\mathrm{we}\mathrm{get}5\mathrm{y}=2\mathrm{x}$$$$\mathrm{substitute}\mathrm{into}\mathrm{eq}\mathrm{of}\mathrm{ellipse}$$$${2\mathrm{x}}^2-{\left(\frac{2\mathrm{x}}{5}\right)}^2=8;{46\mathrm{x}}^2=8\times 25$$$$\mathrm{x}=\pm \frac{10}{\sqrt{23}}\Rightarrow \mathrm{y}=\pm \frac{4}{\sqrt{23}}$$$$\mathrm{Now}\mathrm{equation}\mathrm{of}\mathrm{tangent}\mathrm{line}$$$$\mathrm{the}\mathrm{ellipse}\mathrm{is}5\mathrm{x}-\mathrm{y}=\frac{46}{\sqrt{23}}\mathrm{or}$$$$5\mathrm{x}-\mathrm{y}=2\sqrt{23},\mathrm{so}\mathrm{the}\mathrm{closest}$$$$\mathrm{distance}\mathrm{is}\mathrm{required}\mathrm{equals}\mathrm{to}$$$$\mathrm{d}=\frac{2\sqrt{23}}{\sqrt{26}}$$

Commented by liberty last updated on 23/Jan/21

Terms of Service

Privacy Policy

Contact: info@tinkutara.com