For the ellipse frac{x^2}{18}+frac{y^2}{32}=1 | vedclass

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(A) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{32} = 1$,where $a^2 = 18$ and $b^2 = 32$. Since $b^2 > a^2$,the major axis is along th

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$P(0,8), Q(6,0)$

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(A) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{32} = 1$,where $a^2 = 18$ and $b^2 = 32$. Since $b^2 > a^2$,the major axis is along the $y$-axis. The equation of a tangent with slope $m = -\frac{4}{3}$ is given by $y = mx \pm \sqrt{a^2m^2 + b^2}$. Substituting the values: $y = -\frac{4}{3}x \pm \sqrt{18(-\frac{4}{3})^2 + 32} = -\frac{4}{3}x \pm \sqrt{18(\frac{16}{9}) + 32} = -\frac{4}{3}x \pm \sqrt{32 + 32} = -\frac{4}{3}x \pm 8$. This simplifies to $4x + 3y = \pm 24$. For the tangent $4x + 3y = 24$: The $x$-intercept (where $y=0$) is $Q(6,0)$. The $y$-intercept (where $x=0$) is $P(0,8)$. Thus,the points are $P(0,8)$ and $Q(6,0)$.

For the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1$,if a tangent with slope $-\frac{4}{3}$ intersects the major and minor axes at $P$ and $Q$ respectively. Find $P$ and $Q$.

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