Tangents and normals

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${The length of the sub-tangent to the hyperbola x}^{2} - 4 y^{2} = 4 c o r r e s p o n d i n g t o t h e n o r m a l h a v i n g s l o p e u n i t y i s \frac{1}{\sqrt{k}}, t h e n k i s e q u a l t o$

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Solution

$\begin{array}{l} S l o p e o f n o r m a l i s u n i t y \Rightarrow - \frac{d x}{d y} = 1, \frac{d y}{d x} = - 1 \\ f r o m c u r v e \frac{d y}{d x} = \frac{x}{4 y} = - 1 \\ ∴ x = - 4 y sub in curve 16 y^{2} - 4 y^{2} = 4 \\ y = \pm \frac{1}{\sqrt{3}} ∴ x = \mp \frac{4}{\sqrt{3}} \\ ∴ s u b - \tan g e n t = |y . \frac{d y}{d x}| = \frac{1}{\sqrt{3}} \\ ∴ k = 3 \end{array}$

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Tangents and normals

Remember concepts with our Masterclasses.

The length of the sub-tangent to the hyperbola x2-4y2=4 corresponding to the normal having slope unity is 1k, then k is equal to

Slope of normal is unity​⇒−dxdy=1, dydx=−1​from curve dydx=x4y=−1​∴x=−4y sub in curve 16y2−4y2=4 ​​ y=±13 ∴x= ∓43​∴sub−tangent=y.dydx =13​∴k=3

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${The length of the sub-tangent to the hyperbola x}^{2} - 4 y^{2} = 4 c o r r e s p o n d i n g t o t h e n o r m a l h a v i n g s l o p e u n i t y i s \frac{1}{\sqrt{k}}, t h e n k i s e q u a l t o$