Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at (3, -4), one focus at (4, -4) and one vertex at (5, -4). If $mx-y=4, m>0$ is a tangent to the ellipse E, then the value of $5m^2$ is equal to ________.
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For a shifted ellipse $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, the condition for a line $y=mx+c$ to be a tangent is $(c+mh-k)^2 = a^2m^2+b^2$. Using a coordinate shift ($X=x-h, Y=y-k$) often simplifies the problem.
From the given information, we can determine the parameters of the ellipse.
Center: \( C = (h, k) = (3, -4) \).
Focus: \( S = (4, -4) \).
Vertex: \( V = (5, -4) \).
Since the y-coordinates are the same, the major axis is horizontal.
The distance from the center to a vertex is the semi-major axis length, \( a \).
\( a = \text{distance}(C, V) = \sqrt{(5-3)^2 + (-4 - (-4))^2} = 2 \).
The distance from the center to a focus is \( ae \), where \( e \) is the eccentricity.
\( ae = \text{distance}(C, S) = \sqrt{(4-3)^2 + (-4 - (-4))^2} = 1 \).
We find the eccentricity:
\( e = \frac{ae}{a} = \frac{1}{2} \).
The relationship for the semi-minor axis \( b \) is \( b^2 = a^2(1-e^2) \).
\( b^2 = 2^2 \left(1 - \left(\frac{1}{2}\right)^2\right) = 4 \left(1 - \frac{1}{4}\right) = 4 \cdot \frac{3}{4} = 3 \).
The equation of the ellipse is
\( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).
Hence,
\( \frac{(x-3)^2}{4} + \frac{(y+4)^2}{3} = 1 \).
The line \( y = mx - 4 \) is tangent to this ellipse. We use the condition of tangency.
The condition for a line \( y = MX + C \) to be tangent to the ellipse
\( \frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1 \) is
\( C^2 = a^2M^2 + b^2 \).
First, we shift the origin: let \( X = x - 3 \) and \( Y = y + 4 \).
The line becomes
\( Y - 4 = m(X + 3) - 4 \Rightarrow Y = mX + 3m \).
Here, \( M = m \) and \( C = 3m \). The ellipse is
\( \frac{X^2}{4} + \frac{Y^2}{3} = 1 \).
So, \( a^2 = 4 \) and \( b^2 = 3 \).
Apply the condition of tangency:
\( C^2 = a^2M^2 + b^2 \).
\( (3m)^2 = 4m^2 + 3 \).
\( 9m^2 = 4m^2 + 3 \).
\( 5m^2 = 3 \).
