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Question Number 84510 by 698148290 last updated on 13/Mar/20

Answered by jagoll last updated on 14/Mar/20

$$\mathrm{equation}\mathrm{of}\mathrm{tangent}$$$$\Rightarrow \frac{{\mathrm{x}}_1\mathrm{x}}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}_1\mathrm{y}}{{\mathrm{b}}^2}=1$$$$\Rightarrow \frac{\cos \theta \mathrm{x}}{\mathrm{a}}+\frac{\sin \theta \mathrm{y}}{\mathrm{b}}=1$$$$\mathrm{or}\mathrm{bcos}\theta +\mathrm{asin}\theta =\mathrm{ab}$$

Answered by jagoll last updated on 14/Mar/20

$$\mathrm{equation}\mathrm{of}\mathrm{normal}$$$$\Rightarrow \frac{{\mathrm{x}}_1\mathrm{y}}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}_1\mathrm{x}}{{\mathrm{b}}^2}=1$$$$\Rightarrow \frac{\cos \theta \mathrm{y}}{\mathrm{a}}-\frac{\sin \theta \mathrm{x}}{\mathrm{b}}=1$$$$\mathrm{bcos}\theta \mathrm{y}-\mathrm{asin}\theta \mathrm{x}=\mathrm{ab}$$

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