Let the tangent drawn to the parabola $$y^{2}=24 x$$ at the point $$(\alpha, \beta)$$ is perpendicular to the line $$2 x+2 y=5$$. Then the normal to the hyperbola $$\frac{x^{2}}{\alpha^{2}}-\frac{y^{2}}{\beta^{2}}=1$$ at the point $$(\alpha+4, \beta+4)$$ does NOT pass through the point:

If the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ meets the line $$\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$$ on the $$x$$-axis and the line $$\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$$ on the $$y$$-axis, then the eccentricity of the ellipse is

The tangents at the points $$A(1,3)$$ and $$B(1,-1)$$ on the parabola $$y^{2}-2 x-2 y=1$$ meet at the point $$P$$. Then the area (in unit $${ }^{2}$$ ) of the triangle $$P A B$$ is:

Let the foci of the ellipse $$\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$$ and the hyperbola $$\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}$$ coincide. Then the length of the latus rectum of the hyperbola is: