Question

The least intercept made by a tangent to the ellipse \(\dfrac{x^2}{64}+\dfrac{y^2}{49}=1\) with coordinate axes is

• A. \(40\)
• B. \(10\)
• C. \(15\)
• D. \(100\)

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Write the given ellipse in standard form.
The ellipse is \[ \frac{x^2}{64}+\frac{y^2}{49}=1 \] Comparing with \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] we get \[ a=8,\qquad b=7 \]

Step 2: Write the tangent to the ellipse.
The tangent to the ellipse at parameter \(\theta\) is \[ \frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1 \] Therefore, \[ \frac{x\cos\theta}{8}+\frac{y\sin\theta}{7}=1 \]

Step 3: Find the intercepts on the coordinate axes.
Putting \(y=0\), \[ \frac{x\cos\theta}{8}=1 \] So, \[ x=8\sec\theta \] Putting \(x=0\), \[ \frac{y\sin\theta}{7}=1 \] So, \[ y=7\cosec\theta \] Hence, the length of the intercept between the coordinate axes is \[ L=\sqrt{(8\sec\theta)^2+(7\cosec\theta)^2} \] \[ L^2=64\sec^2\theta+49\cosec^2\theta \]

Step 4: Minimize \(L^2\).
Let \[ u=\tan\theta \] Then, \[ \sec^2\theta=1+\tan^2\theta=1+u^2 \] and \[ \cosec^2\theta=1+\cot^2\theta=1+\frac{1}{u^2} \] Therefore, \[ L^2=64(1+u^2)+49\left(1+\frac{1}{u^2}\right) \] \[ L^2=113+64u^2+\frac{49}{u^2} \] Using \[ A+B\geq 2\sqrt{AB}, \] we get \[ 64u^2+\frac{49}{u^2}\geq 2\sqrt{64u^2\cdot \frac{49}{u^2}} \] \[ 64u^2+\frac{49}{u^2}\geq 2\sqrt{3136} \] \[ 64u^2+\frac{49}{u^2}\geq 112 \] Thus, \[ L^2\geq 113+112 \] \[ L^2\geq 225 \] Therefore, \[ L\geq 15 \]

Step 5: Final conclusion.
Hence, the least intercept made by the tangent with the coordinate axes is \[ \boxed{15} \]