If P(theta) lies on the hyperbola frac{x^{2}} | vedclass

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(B) The parametric coordinates of a point $P$ on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ are $(a \sec 	heta, b 	an 	heta)$.The fo

Vedclass · Accepted Answer

$a^{2} 	an ^{2} 	heta + b^{2} \sec ^{2} 	heta$

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(B) The parametric coordinates of a point $P$ on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ are $(a \sec 	heta, b 	an 	heta)$.The foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$,where $e$ is the eccentricity given by $e^{2} = 1 + \frac{b^{2}}{a^{2}}$,so $a^{2}e^{2} = a^{2} + b^{2}$.The distance $SP = \sqrt{(a \sec 	heta - ae)^{2} + (b 	an 	heta - 0)^{2}} = \sqrt{a^{2}(\sec 	heta - e)^{2} + b^{2} 	an ^{2} 	heta}$.Using $b^{2} = a^{2}(e^{2}-1)$,we get $SP = |ae \sec 	heta - a|$.Similarly,$S^{\prime}P = |ae \sec 	heta + a|$.Therefore,$SP \cdot S^{\prime}P = |(ae \sec 	heta - a)(ae \sec 	heta + a)| = |a^{2}e^{2} \sec ^{2} 	heta - a^{2}|$.Substituting $a^{2}e^{2} = a^{2} + b^{2}$,we get $SP \cdot S^{\prime}P = |(a^{2} + b^{2}) \sec ^{2} 	heta - a^{2}| = |a^{2}(\sec ^{2} 	heta - 1) + b^{2} \sec ^{2} 	heta| = |a^{2} 	an ^{2} 	heta + b^{2} \sec ^{2} 	heta|$.Thus,$SP \cdot S^{\prime}P = a^{2} 	an ^{2} 	heta + b^{2} \sec ^{2} 	heta$.

If $P(\theta)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $S$ and $S^{\prime}$ are the foci of the hyperbola,then $SP \cdot S^{\prime}P =$

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