Let E be an ellipse whose axes are parallel t | vedclass

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(C) The center of the ellipse is $C(3, -4)$.The focus is $S(4, -4)$ and the vertex is $A(5, -4)$.Since the $y$-coordinates are the same,the major axis

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$3$

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(C) The center of the ellipse is $C(3, -4)$.The focus is $S(4, -4)$ and the vertex is $A(5, -4)$.Since the $y$-coordinates are the same,the major axis is horizontal.The distance from the center to the vertex is $a = |5 - 3| = 2$.The distance from the center to the focus is $ae = |4 - 3| = 1$.Thus,$e = \frac{1}{2}$.Using $b^{2} = a^{2}(1 - e^{2})$,we get $b^{2} = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.The equation of the ellipse is $\frac{(x - 3)^{2}}{4} + \frac{(y + 4)^{2}}{3} = 1$.The condition for the line $y = mx - 4$ to be a tangent to the ellipse $\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1$ is $(y - k) = m(x - h) \pm \sqrt{a^{2}m^{2} + b^{2}}$.Here,$h = 3, k = -4, a^{2} = 4, b^{2} = 3$.The line is $y + 4 = mx - 3m$,so $y - (-4) = m(x - 3) - 3m$.Comparing with the tangent form,the constant term is $-3m = \pm \sqrt{4m^{2} + 3}$.Squaring both sides,$9m^{2} = 4m^{2} + 3$.$5m^{2} = 3$.

Let $E$ be an ellipse whose axes are parallel to the coordinate axes,having its center at $(3, -4)$,one focus at $(4, -4)$,and one vertex at $(5, -4)$. If $mx - y = 4$ with $m > 0$ is a tangent to the ellipse $E$,then the value of $5m^{2}$ is equal to .....

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