Let the hyperbola H : frac{x^{2}}{a^{2}}-frac | vedclass

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(B) The hyperbola is $H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. The foci are $S(ae, 0)$ and $S'(-ae, 0)$.For the parabola,the focus is $(ae, 0)$

Vedclass · Accepted Answer

$(3\sqrt{3}, -6\sqrt{2})$

Vedclass · Answer

(B) The hyperbola is $H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. The foci are $S(ae, 0)$ and $S'(-ae, 0)$.For the parabola,the focus is $(ae, 0)$ and the directrix is $x = -ae$.The distance between the focus and the directrix is $2ae$. Since the distance between the focus and directrix of a parabola is $2p$ (where $4p$ is the latus rectum),we have $2p = 2ae$,so $p = ae$.The length of the latus rectum of the parabola is $4p = 4ae$.The length of the latus rectum of $H$ is $\frac{2b^{2}}{a}$.Given $4ae = e 	imes \frac{2b^{2}}{a}$,we get $4a = \frac{2b^{2}}{a}$,so $b^{2} = 2a^{2}$.Since $(2\sqrt{2}, -2\sqrt{2})$ lies on $H$,$\frac{8}{a^{2}} - \frac{8}{b^{2}} = 1$. Substituting $b^{2} = 2a^{2}$,we get $\frac{8}{a^{2}} - \frac{4}{a^{2}} = 1$,so $a^{2} = 4$ and $b^{2} = 8$.Then $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + 2 = 3$,so $e = \sqrt{3}$.The focus of the parabola is $(ae, 0) = (2\sqrt{3}, 0)$ and the directrix is $x = -2\sqrt{3}$.The equation of the parabola is $(y-0)^{2} = 4(ae)(x - (-ae)) = 4(2\sqrt{3})(x + 2\sqrt{3})$ is incorrect based on standard form. Using $y^{2} = 4p(x - h)$,where $p = ae = 2\sqrt{3}$ and vertex is at origin relative to focus/directrix,the parabola is $y^{2} = 4(2\sqrt{3})x = 8\sqrt{3}x$.Testing points: For $(3\sqrt{3}, -6\sqrt{2})$,$y^{2} = (-6\sqrt{2})^{2} = 72$ and $8\sqrt{3}x = 8\sqrt{3}(3\sqrt{3}) = 72$. Thus,$(3\sqrt{3}, -6\sqrt{2})$ lies on the parabola.

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