Question Number 46084 by peter frank last updated on 20/Oct/18
Commented by peter frank last updated on 21/Oct/18
$$\mathrm{help}\: \\ $$
Answered by ajfour last updated on 21/Oct/18
![A≡[acos (α+β), bsin (α+β)] B≡[acos (α−β), bsin (α−β)] slope of chord _(−) = (b/a)(((cos αsin β)/(−sin αsin β))) = −(b/a)cot α (slope of tangent)∣_(θ=α) = (dy/dx)∣_(θ=α) = (((dy/dθ)/(dx/dθ)))∣_(θ=α) = −((bcos α)/(asin α)) .](Q46133.png)
$${A}\equiv\left[{a}\mathrm{cos}\:\left(\alpha+\beta\right),\:{b}\mathrm{sin}\:\left(\alpha+\beta\right)\right] \\ $$$${B}\equiv\left[{a}\mathrm{cos}\:\left(\alpha−\beta\right),\:{b}\mathrm{sin}\:\left(\alpha−\beta\right)\right] \\ $$$$\underset{−} {{slope}\:{of}\:{chord}\:} \\ $$$$\:\:\:=\:\frac{{b}}{{a}}\left(\frac{\mathrm{cos}\:\alpha\mathrm{sin}\:\beta}{−\mathrm{sin}\:\alpha\mathrm{sin}\:\beta}\right)\:=\:−\frac{{b}}{{a}}\mathrm{cot}\:\alpha \\ $$$$\left({slope}\:{of}\:{tangent}\right)\mid_{\theta=\alpha} \:=\:\frac{{dy}}{{dx}}\mid_{\theta=\alpha} \\ $$$$\:\:\:\:\:\:\:\:=\:\left(\frac{{dy}/{d}\theta}{{dx}/{d}\theta}\right)\mid_{\theta=\alpha} \:=\:−\frac{{b}\mathrm{cos}\:\alpha}{{a}\mathrm{sin}\:\alpha}\:. \\ $$
Commented by peter frank last updated on 21/Oct/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$