If $m_1$ and $m_2$ are the slopes of the tangents to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ which passes through $(5,4)$, then the value of $(m_1 + m_2) - (m_1m_2)$ is equal to
Step-by-Step Solution
Key Concept: For a tangent to an ellipse from an external point, substitute the point condition into the tangent equation to find the slope
The tangent passes through $(5,4)$, so $4 = 5m ± \sqrt{10m^2 + 9}$. Rearranging: $4 - 5m = ±\sqrt{10m^2 + 9}$. Squaring: $(4-5m)^2 = 10m^2 + 9$, which gives $16 - 40m + 25m^2 = 10m^2 + 9$, simplifying to $15m^2 - 40m + 7 = 0$. Using the quadratic formula: $m_1 + m_2 = \frac{40}{15} = \frac{8}{3}$, and $m_1m_2 = \frac{7}{15}$. Therefore $(m_1 + m_2) - (m_1m_2) = \frac{8}{3} - \frac{7}{15} = \frac{33}{15}$.
Correct Answer: 3