Question

The line $x \cos a + y \sin a=p$ touches the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ if
A. $a^{2} \cos ^{2} a - b^{2} \sin ^{2} a = p^{2}$
B. $a^{2} \cos ^{2} a - b^{2} \sin ^{2} a = p$
C. $a^{2} \cos ^{2} a - b^{2} \cos ^{2} a = 2p$
D. $a^{2} \cos ^{2} a + b^{2} \sin ^{2} a = p$

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Answer


The correct option is A

[ a^{2} \cos^2(a) - b^{2} \sin^2(a) = p^{2} ]

Given a line of the form ( y = mx + c ), it touches the hyperbola when,

[ c = \sqrt{a^{2} m^{2} - b^{2}} ]

For the given line ( x \cos(a) + y \sin(a) = p ), we can rewrite it in the form ( y = mx + c ) as follows:

[ y = -x \cot(a) + p \csc(a) ]

From the tangent condition and substituting (|m| = \cot(a)) and (c = p \csc(a)), we get:

[ p \csc(a) = \sqrt{a^{2} \cot^{2}(a) - b^{2}} ]

Squaring both sides of the equation, we have:

[ p^{2} \csc^{2}(a) = a^{2} \frac{\cos^{2}(a)}{\sin^{2}(a)} - b^{2} ]

So, simplifying this, we find:

[ p^{2} = a^{2} \cos^{2}(a) - b^{2} \sin^{2}(a) ]

Thus, the line touches the hyperbola if:

[ a^{2} \cos^{2}(a) - b^{2} \sin^{2}(a) = p^{2} ]

Therefore, the correct option is A.


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