The foci of a hyperbola are (pm 2, 0) and its | vedclass

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(B) The equation of the hyperbola is $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$.Given foci $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$. With $e = \frac{3}{2}$

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The equation of the hyperbola is $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$.Given foci $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$. With $e = \frac{3}{2}$,we get $a = \frac{4}{3}$.Using $B^2 = A^2(e^2 - 1)$,we have $B^2 = \frac{16}{9}(\frac{9}{4} - 1) = \frac{16}{9} \cdot \frac{5}{4} = \frac{20}{9}$.The slope of the line $2x + 3y = 6$ is $-\frac{2}{3}$. The slope of the tangent perpendicular to this line is $m = \frac{3}{2}$.The equation of the tangent is $y = mx \pm \sqrt{A^2m^2 - B^2}$.$y = \frac{3}{2}x \pm \sqrt{\frac{16}{9} \cdot \frac{9}{4} - \frac{20}{9}} = \frac{3}{2}x \pm \sqrt{4 - \frac{20}{9}} = \frac{3}{2}x \pm \sqrt{\frac{16}{9}} = \frac{3}{2}x \pm \frac{4}{3}$.Since the point is in the first quadrant,we choose the negative sign for the intercept: $y = \frac{3}{2}x - \frac{4}{3}$.For $x$-intercept $a$,set $y=0$: $0 = \frac{3}{2}a - \frac{4}{3} \Rightarrow a = \frac{8}{9}$.For $y$-intercept $b$,set $x=0$: $b = -\frac{4}{3}$.Thus,$|6a| + |5b| = |6(\frac{8}{9})| + |5(-\frac{4}{3})| = \frac{16}{3} + \frac{20}{3} = \frac{36}{3} = 12$.

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