Category: Coordinate Geometry

Question Number 2158 by Yozzis last updated on 05/Nov/15 $${Find}\:{the}\:{ratio},\:{over}\:{one}\:{revolution},\:{of}\:{the}\:{distance}\:{moved}\:{by} \\ $$$${a}\:{wheel}\:{rolling}\:{on}\:{a}\:{flat}\:{surface}\:{to}\:{the}\:{distance}\:{traced}\:{out}\:{by} \\ $$$${a}\:{point}\:{on}\:{its}\:{circumference}.\: \\ $$ Commented by ssahoo last updated on 06/Nov/15 $$\mathrm{Wheel}\:\mathrm{distance}=\mathrm{2}\pi{r} \\…

Question Number 132433 by john_santu last updated on 14/Feb/21 Answered by liberty last updated on 14/Feb/21 $$\mathrm{by}\:\mathrm{using}\:\mathrm{Cosine}\:\mathrm{of}\:\mathrm{law}\: \\ $$$$\:\left(\mathrm{r}−\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0}.\mathrm{5}^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}−\mathrm{r}\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{0}.\mathrm{5}×\left(\sqrt{\mathrm{2}}−\mathrm{r}\right)\mathrm{cos}\:\mathrm{45}° \\ $$$$\cancel{\mathrm{r}^{\mathrm{2}} }−\mathrm{r}+\cancel{\frac{\mathrm{1}}{\mathrm{4}}}=\cancel{\frac{\mathrm{1}}{\mathrm{4}}}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{r}+\cancel{\mathrm{r}^{\mathrm{2}}…

Question Number 132131 by Ar Brandon last updated on 11/Feb/21 Commented by Ar Brandon last updated on 11/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{the}\:\mathrm{areas}\:\mathrm{of}\: \\ $$$$\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{and}\:\mathrm{triangle}\:\mathrm{A}'\mathrm{B}'\mathrm{C}' \\ $$ Commented by…

Question Number 66263 by peter frank last updated on 12/Aug/19 Commented by peter frank last updated on 12/Aug/19 $${please}\:{help} \\ $$ Terms of Service Privacy…

Question Number 66161 by peter frank last updated on 09/Aug/19 Answered by mr W last updated on 10/Aug/19 $${if}\:{a}\:{line}\:{y}={mx}+{c}\:{tangents}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}…