If 3x + 4y = 12√2 is a tangent to the ellipse x²/a² + y²/9 = 1 for some a ∈ R, then the distance between foci of the ellipse is:
2√7
The line lx + my = n is tangent to the ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) if \(n^2 = a^2 l^2 + b^2 m^2\).
Here l = 3, m = 4, n = 12√2, b² = 9. Substituting: \((12\sqrt2)^2 = a^2(3)^2 + 9(4)^2\).
This gives \(288 = 9a^2 + 144 \Rightarrow 9a^2 = 144 \Rightarrow a^2 = 16\).
Since a² = 16 > b² = 9, the major axis is along the x-axis, and \(c^2 = a^2-b^2 = 16-9 = 7\), so c = √7.
The distance between the foci is \(2c = 2\sqrt7\).
Similar Questions
Let $\frac{x²}{a²}$ +$\frac{y²}{b²}$ = 1 (a > b) be a given ellipse. Length of its Latus rectum is 10. If its eccentricity is the maximum value of the function φ(t) = $\frac{5}{12}$ + t − t², then a² + b² is equal to:
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