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Question-84510




Question Number 84510 by 698148290 last updated on 13/Mar/20
Answered by jagoll last updated on 14/Mar/20
equation of tangent  ⇒ ((x_1 x)/a^2 ) + ((y_1 y)/b^2 ) = 1  ⇒ ((cos θ x)/a) + ((sin θ y)/b) = 1   or bcos θ + asin θ = ab
$$\mathrm{equation}\:\mathrm{of}\:\mathrm{tangent} \\ $$$$\Rightarrow\:\frac{\mathrm{x}_{\mathrm{1}} \mathrm{x}}{\mathrm{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{y}_{\mathrm{1}} \mathrm{y}}{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\frac{\mathrm{cos}\:\theta\:\mathrm{x}}{\mathrm{a}}\:+\:\frac{\mathrm{sin}\:\theta\:\mathrm{y}}{\mathrm{b}}\:=\:\mathrm{1}\: \\ $$$$\mathrm{or}\:\mathrm{bcos}\:\theta\:+\:\mathrm{asin}\:\theta\:=\:\mathrm{ab} \\ $$
Answered by jagoll last updated on 14/Mar/20
equation of normal  ⇒ ((x_1 y)/a^2 ) − ((y_1 x)/b^2 ) = 1  ⇒ ((cos θ y)/a) − ((sin θ x)/b) = 1  bcos θ y − asin θ x = ab
$$\mathrm{equation}\:\mathrm{of}\:\mathrm{normal} \\ $$$$\Rightarrow\:\frac{\mathrm{x}_{\mathrm{1}} \mathrm{y}}{\mathrm{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{y}_{\mathrm{1}} \mathrm{x}}{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\frac{\mathrm{cos}\:\theta\:\mathrm{y}}{\mathrm{a}}\:−\:\frac{\mathrm{sin}\:\theta\:\mathrm{x}}{\mathrm{b}}\:=\:\mathrm{1} \\ $$$$\mathrm{bcos}\:\theta\:\mathrm{y}\:−\:\mathrm{asin}\:\theta\:\mathrm{x}\:=\:\mathrm{ab} \\ $$

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