Menu Close

Category: Coordinate Geometry

prove-that-product-of-lengths-of-perpendiculars-from-any-point-of-hyperbola-to-its-asymptotes-is-constant-

Question Number 43921 by peter frank last updated on 17/Sep/18 $${prove}\:{that}\:{product}\:{of}\:{lengths}\:{of}\:{perpendiculars} \\ $$$${from}\:{any}\:{point}\:{of}\:{hyperbola}\:{to}\:{its} \\ $$$${asymptotes}\:{is}\:{constant} \\ $$ Answered by math1967 last updated on 18/Sep/18 $${let}\:{equn}.\:{of}\:{hyperbola}\:{is}\frac{{x}^{\mathrm{2}}…

The-tangent-at-P-to-an-ellipse-meet-diretrix-at-Q-prove-that-the-line-joining-the-corresponding-focus-to-P-and-Q-are-perpendicular-

Question Number 43919 by peter frank last updated on 17/Sep/18 $${The}\:{tangent}\:{at}\:{P}\:{to}\:{an}\:{ellipse}\:{meet}\: \\ $$$${diretrix}\:{at}\:{Q}.{prove}\:{that}\:{the}\:{line}\:{joining} \\ $$$${the}\:{corresponding}\:{focus}\:{to}\:{P}\:{and}\:{Q}\:{are}\:{perpendicular} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

if-tan-x-y-3-4-and-tan-x-y-8-15-show-that-tan-2x-77-36-

Question Number 43875 by peter frank last updated on 16/Sep/18 $${if}\:\:\:\mathrm{tan}\:\left({x}+{y}\right)=\frac{\mathrm{3}}{\mathrm{4}}\:\:{and}\:\mathrm{tan}\left({x}−{y}\right)=\frac{\mathrm{8}}{\mathrm{15}}\: \\ $$$${show}\:{that}\:\:\mathrm{tan}\:\mathrm{2}{x}=\:\frac{\mathrm{77}}{\mathrm{36}} \\ $$ Answered by ajfour last updated on 16/Sep/18 $$\mathrm{tan}\:\left({x}+{y}+{x}−{y}\right)=\frac{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{8}}{\mathrm{15}}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{5}}} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{tan}\:\mathrm{2}{x}\:=\:\frac{\mathrm{77}}{\mathrm{36}}\:.…

Question-43404

Question Number 43404 by peter frank last updated on 10/Sep/18 Answered by alex041103 last updated on 10/Sep/18 $$\frac{{d}}{{dx}}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{16}}−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}\right) \\ $$$$\frac{{x}}{\mathrm{8}}−\frac{{y}}{\mathrm{2}}\left(\frac{{dy}}{{dx}}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{x}}{\mathrm{4}{y}} \\…

Question-43384

Question Number 43384 by peter frank last updated on 10/Sep/18 Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18 $${cos}\mathrm{30}^{{o}} =\frac{\frac{{a}}{\mathrm{2}}}{{R}} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{{a}}{\mathrm{2}{R}} \\ $$$${a}={R}\sqrt{\mathrm{3}}\: \\ $$$${a}=\mathrm{5}\sqrt{\mathrm{3}}\:…