Step 1: Understanding the Concept:
A tangent to the ellipse is a line. This line acts as a chord for the circle \( x^2 + y^2 = 25 \). The length of a chord of a circle with radius \( R \) is given by \( L = 2\sqrt{R^2 - d^2} \), where \( d \) is the perpendicular distance from the center of the circle (origin) to the line.
Step 2: Key Formula or Approach:
1. Equation of tangent to \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( y = mx \pm \sqrt{a^2 m^2 + b^2} \).
2. Perpendicular distance from \( (0,0) \) is \( d = \frac{\sqrt{a^2 m^2 + b^2}}{\sqrt{1 + m^2}} \).
3. Minimize chord length \( L = 2\sqrt{25 - d^2} \), which is equivalent to maximizing \( d^2 \).
Step 3: Detailed Explanation:
For the ellipse \( a^2 = 4 \) and \( b^2 = 9 \). The tangent is \( mx - y + \sqrt{4m^2 + 9} = 0 \).
Distance \( d \) from origin:
\[ d^2 = \frac{4m^2 + 9}{1 + m^2} \]
Let's analyze \( d^2 \) as a function of \( m^2 \). Let \( u = m^2 \):
\[ d^2(u) = \frac{4u + 9}{u + 1} = \frac{4(u+1) + 5}{u+1} = 4 + \frac{5}{u+1} \]
To minimize the intercept length \( L = 2\sqrt{25 - d^2} \), we need to maximize \( d^2 \).
The function \( 4 + \frac{5}{u+1} \) is maximized when the denominator \( u+1 \) is minimized.
Since \( u = m^2 \ge 0 \), the minimum value of \( u \) is 0.
Maximum \( d^2 = 4 + \frac{5}{0+1} = 9 \).
(This happens when \( m = 0 \), i.e., the tangent is horizontal).
Minimum intercept length \( L = 2\sqrt{25 - 9} = 2\sqrt{16} = 2 \cdot 4 = 8 \).
Step 4: Final Answer:
The minimum length of the intercept is 8.