If P(x_1, y_1) is a point on the hyperbola x^ | vedclass

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(B) The given hyperbola is $x^2 - y^2 = a^2$,which is a rectangular hyperbola with eccentricity $e = \sqrt{2}$.The foci are $S(ae, 0) = (a\sqrt{2}, 0)

Vedclass · Accepted Answer

$x_1^2 + y_1^2$

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(B) The given hyperbola is $x^2 - y^2 = a^2$,which is a rectangular hyperbola with eccentricity $e = \sqrt{2}$.The foci are $S(ae, 0) = (a\sqrt{2}, 0)$ and $S'(-ae, 0) = (-a\sqrt{2}, 0)$.For any point $P(x_1, y_1)$ on the hyperbola,the focal distances are given by $SP = |ex_1 - a|$ and $S'P = |ex_1 + a|$.Since $e = \sqrt{2}$,we have $SP = |\sqrt{2}x_1 - a|$ and $S'P = |\sqrt{2}x_1 + a|$.Therefore,$SP \cdot S'P = |(\sqrt{2}x_1 - a)(\sqrt{2}x_1 + a)| = |2x_1^2 - a^2|$.Since $P(x_1, y_1)$ lies on $x^2 - y^2 = a^2$,we have $x_1^2 - y_1^2 = a^2$,which implies $x_1^2 = a^2 + y_1^2$.Substituting this into the expression: $SP \cdot S'P = |2(a^2 + y_1^2) - a^2| = |a^2 + 2y_1^2|$.However,for a rectangular hyperbola,the product of focal distances $SP \cdot S'P$ is equal to the square of the distance from the center to the point $P$,which is $x_1^2 + y_1^2$.

If $P(x_1, y_1)$ is a point on the hyperbola $x^2 - y^2 = a^2$,then $SP \cdot S'P = \_\_\_\_$

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