Question Number 84510 by 698148290 last updated on 13/Mar/20
Answered by jagoll last updated on 14/Mar/20
$$\mathrm{equation}\:\mathrm{of}\:\mathrm{tangent} \\ $$$$\Rightarrow\:\frac{\mathrm{x}_{\mathrm{1}} \mathrm{x}}{\mathrm{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{y}_{\mathrm{1}} \mathrm{y}}{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\frac{\mathrm{cos}\:\theta\:\mathrm{x}}{\mathrm{a}}\:+\:\frac{\mathrm{sin}\:\theta\:\mathrm{y}}{\mathrm{b}}\:=\:\mathrm{1}\: \\ $$$$\mathrm{or}\:\mathrm{bcos}\:\theta\:+\:\mathrm{asin}\:\theta\:=\:\mathrm{ab} \\ $$
Answered by jagoll last updated on 14/Mar/20
$$\mathrm{equation}\:\mathrm{of}\:\mathrm{normal} \\ $$$$\Rightarrow\:\frac{\mathrm{x}_{\mathrm{1}} \mathrm{y}}{\mathrm{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{y}_{\mathrm{1}} \mathrm{x}}{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\frac{\mathrm{cos}\:\theta\:\mathrm{y}}{\mathrm{a}}\:−\:\frac{\mathrm{sin}\:\theta\:\mathrm{x}}{\mathrm{b}}\:=\:\mathrm{1} \\ $$$$\mathrm{bcos}\:\theta\:\mathrm{y}\:−\:\mathrm{asin}\:\theta\:\mathrm{x}\:=\:\mathrm{ab} \\ $$