What is the distance between the directrices | vedclass

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(B) The given equations are $x = 8 \sec 	heta$ and $y = 8 	an 	heta$.Squaring and subtracting, we get $x^2 - y^2 = 64 \sec^2 	heta - 64 	an^2 	h

Vedclass · Accepted Answer

$8$

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(B) The given equations are $x = 8 \sec 	heta$ and $y = 8 	an 	heta$.Squaring and subtracting, we get $x^2 - y^2 = 64 \sec^2 	heta - 64 	an^2 	heta = 64(\sec^2 	heta - 	an^2 	heta) = 64$.This is a rectangular hyperbola of the form $x^2 - y^2 = a^2$, where $a = 8$.For a rectangular hyperbola, the eccentricity $e = \sqrt{2}$.The distance between the directrices of a hyperbola is given by $\frac{2a}{e}$.Substituting the values, the distance $= \frac{2(8)}{\sqrt{2}} = \frac{16}{\sqrt{2}} = 8 \sqrt{2}$.

What is the distance between the directrices of the hyperbola $x = 8 \sec \theta, y = 8 \tan \theta$ (in $\sqrt{2}$)?

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