Question:
Let \(L\) be a common tangent line to the curves
\(4x^2+9y^2=36\) and \((2x)^2+(2y)^2=31\). Then the square of the slope of the line \(L\) is __________.
Let \(L\) be a common tangent line to the curves
\(4x^2+9y^2=36\) and \((2x)^2+(2y)^2=31\). Then the square of the slope of the line \(L\) is __________.
\(4x^2+9y^2=36\) and \((2x)^2+(2y)^2=31\). Then the square of the slope of the line \(L\) is __________.
Show Hint
Tangent \(y=mx+c\) to the ellipse needs \(c^2=9m^2+4\) and to the circle \(x^2+y^2=\frac{31}{4}\) needs \(c^2=\frac{31}{4}(1+m^2)\).
Updated On: Nov 4, 2026
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Correct Answer: 3
Solution and Explanation
Ellipse: \(\frac{x^2}{9}+\frac{y^2}{4}=1\), tangent \(y=mx+c\) with \(c^2=9m^2+4\).
Circle: \(x^2+y^2=\frac{31}{4}\), tangent condition \(c^2=\frac{31}{4}(1+m^2)\).
Equate: \(9m^2+4=\frac{31}{4}+\frac{31}{4}m^2\Rightarrow36m^2+16=31+31m^2\Rightarrow5m^2=15\Rightarrow m^2=3\).
Answer: 3
Circle: \(x^2+y^2=\frac{31}{4}\), tangent condition \(c^2=\frac{31}{4}(1+m^2)\).
Equate: \(9m^2+4=\frac{31}{4}+\frac{31}{4}m^2\Rightarrow36m^2+16=31+31m^2\Rightarrow5m^2=15\Rightarrow m^2=3\).
Answer: 3
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