The least intercept made by a tangent to the ellipse \(\dfrac{x^2}{64}+\dfrac{y^2}{49}=1\) with coordinate axes is
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For an ellipse
\[
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,
\]
the tangent
\[
\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1
\]
cuts intercepts \(a\sec\theta\) and \(b\cosec\theta\) on the coordinate axes.
Step 1: Write the ellipse in standard form and identify a and b. x²/64 + y²/49 = 1 → a = 8, b = 7. Step 2: Write the tangent line in parametric form. The tangent at parameter θ is (x cos θ)/8 + (y sin θ)/7 = 1. Step 3: Find the intercepts on the axes. x-intercept (y = 0): x = 8 sec θ. y-intercept (x = 0): y = 7 cosec θ. The intercept length squared is L² = 64 sec²θ + 49 cosec²θ. Step 4: Express in terms of u = tan θ and minimize. L² = 64(1 + u²) + 49(1 + 1/u²) = 113 + 64u² + 49/u². By AM-GM, 64u² + 49/u² ≥ 2√(64·49) = 112. Hence L² ≥ 225 → L ≥ 15. Step 5: Final conclusion. The least intercept length is 15.