Question

If a tangent of slope 2 to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) touches the circle \(x^2 + y^2 = 4\), then the maximum value of ab is:

• A. \(4\)
• B. \(12\)
• C. \(5\)
• D. \(7\)

Hint:

Use the condition that the perpendicular distance from the center of the circle to the tangent equals the radius. Use AM-GM inequality to maximize ab.

Answer 1

The Correct Option is C

 

Step 1: Write the equation of the tangent to the ellipse with slope 2.
The equation of a tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with slope m is given by:
\(y = mx \pm \sqrt{a^2m^2 + b^2}\). 
Given slope m = 2, the equation of the tangent is:
\(y = 2x \pm \sqrt{4a^2 + b^2}\). 

Step 2: Use the condition that the tangent touches the circle x² + y² = 4.
The perpendicular distance from the center (0, 0) of the circle to the tangent must be equal to the radius, which is 2. 
The equation of the tangent can be written as \(2x - y \pm \sqrt{4a^2 + b^2} = 0\). 
The perpendicular distance is:
\(\frac{|2(0) - 0 \pm \sqrt{4a^2 + b^2}|}{\sqrt{2^2 + (-1)^2}} = 2\) 
\(\frac{\sqrt{4a^2 + b^2}}{\sqrt{5}} = 2\) 
\(\sqrt{4a^2 + b^2} = 2\sqrt{5}\) 
\(4a^2 + b^2 = 20\). 

Step 3: Find the maximum value of ab.
We want to maximize ab. 
By AM-GM inequality, \(\frac{4a^2 + b^2}{2} \ge \sqrt{4a^2b^2} = 2ab\). 
\(4a^2 + b^2 \ge 4ab\). 
Since \(4a^2 + b^2 = 20\), we have \(20 \ge 4ab\). 
\(ab \le 5\). 
The maximum value of ab is 5. 

Step 4: Verify the equality condition.
Equality holds when \(4a^2 = b^2\). 
Substituting in \(4a^2 + b^2 = 20\), we get \(2b^2 = 20\), so \(b^2 = 10\) and \(b = \sqrt{10}\). 
\(4a^2 = 10\), so \(a^2 = \frac{10}{4} = \frac{5}{2}\) and \(a = \sqrt{\frac{5}{2}}\). 
Then \(ab = \sqrt{\frac{5}{2}} \times \sqrt{10} = \sqrt{25} = 5\). Therefore, the maximum value of ab is 5.