If the line x cos alpha + y sin alpha = p is | vedclass

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Question

(A) The equation of the line is $y = -(\cot \alpha) x + p \csc \alpha$. Comparing this with the tangent form $y = mx \pm \sqrt{a^2 m^2 - b^2}$,where $

Vedclass · Accepted Answer

$a^2 \cos^2 \alpha - b^2 \sin^2 \alpha = p^2$

Vedclass · Answer

(A) The equation of the line is $y = -(\cot \alpha) x + p \csc \alpha$. Comparing this with the tangent form $y = mx \pm \sqrt{a^2 m^2 - b^2}$,where $m = -\cot \alpha$ and $c = p \csc \alpha$. Squaring both sides,we get $c^2 = a^2 m^2 - b^2$. Substituting the values,$(p \csc \alpha)^2 = a^2 (-\cot \alpha)^2 - b^2$. $p^2 \csc^2 \alpha = a^2 \cot^2 \alpha - b^2$. Multiplying by $\sin^2 \alpha$,we get $p^2 = a^2 \cos^2 \alpha - b^2 \sin^2 \alpha$.

If the line $x \cos \alpha + y \sin \alpha = p$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,then:

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