1. The problem asks to determine the direction in which the parabola $x=4y^2$ opens. 2. Rewrite the equation: $x=4y^2$ can be compared to the standard form of a parabola that opens either left or right, $x=ay^2$. 3. Since the coefficient of $y^2$ is positive ($4$), the parabola opens to the right. Answer: (b) opens right. 1. Find the gradient of the normal to the parabola given by $y^2=4ax$ at point $(at^2, 2at)$. 2. Differentiate implicitly: $2y rac{dy}{dx} = 4a$ which gives $rac{dy}{dx} = rac{2a}{y}$. 3. Substitute point $(at^2, 2at)$: slope of tangent is $rac{dy}{dx} = rac{2a}{2at} = rac{1}{t}$. 4. Gradient of normal is negative reciprocal: $-rac{1}{rac{1}{t}} = -t$. Answer: (c) $-t_1$. 1. Find $c$ so that line $y = 4x + c$ is tangent to ellipse $rac{x^2}{8} + rac{y^2}{4} = 1$. 2. Substitute $y$ from the line into the ellipse: $$ \frac{x^2}{8} + \frac{(4x + c)^2}{4} = 1 $$ 3. Multiply both sides to clear denominators: $$ \frac{x^2}{8} + \frac{16x^2 + 8cx + c^2}{4} = 1 $$ 4. Multiply everything by 8: $$ x^2 + 2(16x^2 + 8cx + c^2) = 8 $$ $$ x^2 + 32x^2 + 16cx + 2c^2 = 8 $$ $$ 33x^2 + 16cx + 2c^2 - 8 = 0 $$ 5. For tangency, discriminant $D=0$: $$ (16c)^2 - 4 \times 33 \times (2c^2 - 8) = 0 $$ $$ 256 c^2 - 132 (2c^2 - 8) = 0 $$ $$ 256 c^2 - 264 c^2 + 1056 = 0 $$ $$ -8 c^2 + 1056 = 0 $$ $$ c^2 = \frac{1056}{8} = 132 $$ 6. Thus, $c = \pm \sqrt{132} = \pm 2 \sqrt{33}$, approximate as $\pm 11.49$. Closest option given: (d) $\pm \sqrt{32}$ is not correct but $\pm 6$ is incorrect as well. The correct exact answer is $\pm 2 \sqrt{33}$. Shortcut summary: 1. For parabola $x=4y^2$, check if coefficient is positive (opens right) or negative (opens left). 2. For gradient of normal, find slope of tangent by implicit differentiation and take negative reciprocal. 3. For tangency of line and ellipse, substitute line into ellipse, obtain quadratic in $x$, set discriminant to zero and solve for $c$.