Let $$\alpha, \beta$$ be the roots of the equation $$x^2+2 \sqrt{2} x-1=0$$. The quadratic equation, whose roots are $$\alpha^4+\beta^4$$ and $$\frac{1}{10}(\alpha^6+\beta^6)$$, is:

Let $$f(x)=x^2+9, g(x)=\frac{x}{x-9}$$ and $$\mathrm{a}=f \circ g(10), \mathrm{b}=g \circ f(3)$$. If $$\mathrm{e}$$ and $$l$$ denote the eccentricity and the length of the latus rectum of the ellipse $$\frac{x^2}{\mathrm{a}}+\frac{y^2}{\mathrm{~b}}=1$$, then $$8 \mathrm{e}^2+l^2$$ is equal to.

The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:

A variable line $$\mathrm{L}$$ passes through the point $$(3,5)$$ and intersects the positive coordinate axes at the points $$\mathrm{A}$$ and $$\mathrm{B}$$. The minimum area of the triangle $$\mathrm{OAB}$$, where $$\mathrm{O}$$ is the origin, is :