1. The equation given is $$-3x^2 + 7y^2 = -1$$. To identify the conic, rewrite it as $$3x^2 - 7y^2 = 1$$ by multiplying both sides by -1. This is of the form $$Ax^2 - By^2 = 1$$ where $$A, B > 0$$, which represents a hyperbola. Answer: Hyperbola. 2. For the hyperbola with foci at $$(\pm 5, 0)$$ and conjugate axis length 6: - The distance between foci is $$2c = 10 \Rightarrow c = 5$$. - The conjugate axis length is $$2b = 6 \Rightarrow b = 3$$. - The relationship $$c^2 = a^2 + b^2$$ gives $$a^2 = c^2 - b^2 = 25 - 9 = 16$$. - Since foci are on the x-axis, the hyperbola equation is $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$. - Substitute $$a^2 = 16$$ and $$b^2 = 9$$: $$\frac{x^2}{16} - \frac{y^2}{9} = 1$$. Answer: Option C. 3. Rotation of axes by 45°: The new coordinates $$(x', y')$$ after rotation by angle $$\theta = 45^\circ$$ are given by: $$x' = x \cos 45^\circ + y \sin 45^\circ$$ $$y' = -x \sin 45^\circ + y \cos 45^\circ$$ Since $$\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$$, For point $$(6,7)$$: $$x' = 6 \times \frac{\sqrt{2}}{2} + 7 \times \frac{\sqrt{2}}{2} = \frac{6+7}{\sqrt{2}} = \frac{13\sqrt{2}}{2}$$ $$y' = -6 \times \frac{\sqrt{2}}{2} + 7 \times \frac{\sqrt{2}}{2} = \frac{-6+7}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ Answer: New coordinates are $$\left( \frac{13\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$$.