The equation of the normal to the hyperbola f | vedclass

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Question

(A) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec 	heta, b 	an 	heta)$ is given by $ax

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$\sqrt{21}(x - 2y) = 41$

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(A) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec 	heta, b 	an 	heta)$ is given by $ax \cos 	heta + by \cot 	heta = a^2 + b^2$.Here,$a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$.The equation of the normal is $5x \cos 	heta + 4y \cot 	heta = 25 + 16 = 41$  .......$(i)$This line is perpendicular to the line $2x + y = 1$,which has a slope of $-2$.The slope of the normal line $(i)$ is $m = -\frac{5 \cos 	heta}{4 \cot 	heta} = -\frac{5 \sin 	heta}{4}$.Since the lines are perpendicular,the product of their slopes is $-1$:$(-\frac{5 \sin 	heta}{4})(-2) = -1 \implies \frac{5 \sin 	heta}{2} = -1 \implies \sin 	heta = -\frac{2}{5}$.Then,$\cos 	heta = \pm \sqrt{1 - \sin^2 	heta} = \pm \sqrt{1 - \frac{4}{25}} = \pm \frac{\sqrt{21}}{5}$.Also,$\cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{\pm \sqrt{21}/5}{-2/5} = \mp \frac{\sqrt{21}}{2}$.Substituting these into equation $(i)$:$5x(\pm \frac{\sqrt{21}}{5}) + 4y(\mp \frac{\sqrt{21}}{2}) = 41$$\pm \sqrt{21}x \mp 2\sqrt{21}y = 41$$\pm \sqrt{21}(x - 2y) = 41$.Taking the positive root,we get $\sqrt{21}(x - 2y) = 41$.

The equation of the normal to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ that is perpendicular to the line $2x + y = 1$ is:

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