1. Problem 27: Find the new coordinates of the point (6,7) after rotating the xy-axes through an angle of 45°. Step 1: The rotation formulas for coordinates when axes are rotated by angle $\theta$ are: $$x' = x\cos\theta + y\sin\theta$$ $$y' = -x\sin\theta + y\cos\theta$$ Step 2: Here, $\theta = 45^\circ$, so $\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}$. Step 3: Substitute $x=6$, $y=7$: $$x' = 6 \times \frac{1}{\sqrt{2}} + 7 \times \frac{1}{\sqrt{2}} = \frac{6+7}{\sqrt{2}} = \frac{13}{\sqrt{2}}$$ $$y' = -6 \times \frac{1}{\sqrt{2}} + 7 \times \frac{1}{\sqrt{2}} = \frac{-6+7}{\sqrt{2}} = \frac{1}{\sqrt{2}}$$ Step 4: So, the new coordinates are $\left(\frac{13}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. Answer: Option A. 2. Problem 28: Find the equation of the normal to $\frac{y^2}{4} - \frac{x^2}{5} = 1$ at point $(\sqrt{5}, 2\sqrt{2})$. Step 1: Differentiate implicitly: $$\frac{2y}{4} \frac{dy}{dx} - \frac{2x}{5} = 0 \implies \frac{y}{2} \frac{dy}{dx} = \frac{2x}{5}$$ Step 2: Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{2x}{5} \times \frac{2}{y} = \frac{4x}{5y}$$ Step 3: Substitute $x=\sqrt{5}$, $y=2\sqrt{2}$: $$\frac{dy}{dx} = \frac{4 \times \sqrt{5}}{5 \times 2\sqrt{2}} = \frac{4\sqrt{5}}{10\sqrt{2}} = \frac{2\sqrt{5}}{5\sqrt{2}}$$ Step 4: Slope of tangent $m_t = \frac{2\sqrt{5}}{5\sqrt{2}}$, so slope of normal $m_n = -\frac{1}{m_t} = -\frac{5\sqrt{2}}{2\sqrt{5}} = -\frac{5}{2} \sqrt{\frac{2}{5}} = -\frac{5}{2} \times \frac{\sqrt{2}}{\sqrt{5}}$. Step 5: Equation of normal line at $(\sqrt{5}, 2\sqrt{2})$: $$y - 2\sqrt{2} = m_n (x - \sqrt{5})$$ Step 6: Multiply both sides to clear radicals and simplify: Multiply both sides by $2\sqrt{5}$: $$2\sqrt{5}(y - 2\sqrt{2}) = -5\sqrt{2}(x - \sqrt{5})$$ Step 7: Expand: $$2\sqrt{5}y - 4\sqrt{10} = -5\sqrt{2}x + 5\sqrt{10}$$ Step 8: Bring all terms to one side: $$5\sqrt{2}x + 2\sqrt{5}y - 9\sqrt{10} = 0$$ Step 9: Divide entire equation by $\sqrt{10}$: $$\sqrt{5}x + \sqrt{2}y - 9 = 0$$ Answer: Option B. 3. Problem 29: Find value of $k$ such that line $y = kx + 1$ is tangent to $3x^2 - 4y^2 = 12$. Step 1: Substitute $y = kx + 1$ into the hyperbola equation: $$3x^2 - 4(kx + 1)^2 = 12$$ Step 2: Expand: $$3x^2 - 4(k^2 x^2 + 2kx + 1) = 12$$ $$3x^2 - 4k^2 x^2 - 8kx - 4 = 12$$ Step 3: Rearrange: $$(3 - 4k^2) x^2 - 8kx - 16 = 0$$ Step 4: For tangency, discriminant $D=0$: $$D = (-8k)^2 - 4(3 - 4k^2)(-16) = 64k^2 + 64(3 - 4k^2) = 64k^2 + 192 - 256k^2 = 192 - 192k^2 = 0$$ Step 5: Solve for $k$: $$192 = 192k^2 \implies k^2 = 1 \implies k = \pm 1$$ Answer: $k = 1$ or $k = -1$.