The lines x cos alpha + y sin alpha = P, alph | vedclass

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(C) The equation of the straight lines joining the origin to the points of intersection of the line $x \cos \alpha + y \sin \alpha = P$ and the hyperb

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$16x^2 + y^2 = 108$

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(C) The equation of the straight lines joining the origin to the points of intersection of the line $x \cos \alpha + y \sin \alpha = P$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{36} = 1$ is obtained by homogenizing the hyperbola equation using the line equation: $\frac{x^2}{9} - \frac{y^2}{36} = (\frac{x \cos \alpha + y \sin \alpha}{P})^2$. Expanding this,we get $x^2(\frac{1}{9} - \frac{\cos^2 \alpha}{P^2}) + y^2(-\frac{1}{36} - \frac{\sin^2 \alpha}{P^2}) - \frac{2xy \cos \alpha \sin \alpha}{P^2} = 0$. Since the lines subtend a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(\frac{1}{9} - \frac{\cos^2 \alpha}{P^2}) + (-\frac{1}{36} - \frac{\sin^2 \alpha}{P^2}) = 0$. This simplifies to $\frac{1}{12} - \frac{1}{P^2} = 0$,so $P^2 = 12$. The polar of a point $(h, k)$ with respect to $\frac{x^2}{9} - \frac{y^2}{36} = 1$ is $\frac{xh}{9} - \frac{yk}{36} = 1$. Comparing this with $x \cos \alpha + y \sin \alpha = P$,we have $\frac{h/9}{\cos \alpha} = \frac{-k/36}{\sin \alpha} = \frac{1}{P}$. Thus,$\cos \alpha = \frac{Ph}{9}$ and $\sin \alpha = -\frac{Pk}{36}$. Using $\cos^2 \alpha + \sin^2 \alpha = 1$,we get $\frac{P^2 h^2}{81} + \frac{P^2 k^2}{1296} = 1$. Substituting $P^2 = 12$,we get $\frac{12h^2}{81} + \frac{12k^2}{1296} = 1$,which simplifies to $\frac{4h^2}{27} + \frac{k^2}{108} = 1$. Multiplying by $108$,we get $16h^2 + k^2 = 108$. Replacing $(h, k)$ with $(x, y)$,the locus is $16x^2 + y^2 = 108$.

The lines $x \cos \alpha + y \sin \alpha = P, \alpha \in R$ are chords of the hyperbola $\frac{x^2}{9} - \frac{y^2}{36} = 1$ and they subtend a right angle at the centre of the hyperbola. The locus of the poles of these lines with respect to the given hyperbola is

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