Calculas Scholarship T2W1

--- title: Calculas Scholarship T2W1 tags: Calculus Scholarship description: Term 2 Week 1 type: pdf slideOptions: theme: white --- ###### tags: `Calculus Scholarship` ### Q1 ### Standard Equation: $(y+1)^2 = 3(x-3)$, $(p, q) = (3, -1)$, $\alpha=\frac{3}{4}$, so we have: - Vertex: $(3,-1)$ - Focus: $(\frac{15}{4}, -1)$ - LR: 3 - Directrix: $x = \frac{9}{4}$ - Eccentricity: 1 ### Q2 ### Based on definition, we have $$|x-(-a)| = \sqrt{(x-a)^2 + y^2}$$ $\implies x^2 + 2ax + a^2 = x^2 - 2ax +a^2 + y^2$ $\implies 4ax = y^2$, as required. ### Q3 ### With E = 0 and A = B, the general quadratic form would be $Ax^2 + Ay^2 + Cx + Dy + F = 0$ $\implies A[x^2+\frac{C}{A}x] + A[y^2+\frac{D}{A}y] + F=0$ $\implies A[(x+\frac{C}{2A})^2-\frac{C^2}{4A^2}] + A[(y+\frac{D}{2A})^2-\frac{D^2}{4A^2}] + F=0$ $\implies A(x+\frac{C}{2A})^2 + A(y+\frac{D}{2A})^2=\frac{C^2+D^2-4AF}{4A}$ $\implies (x+\frac{C}{2A})^2 + (y+\frac{D}{2A})^2=\frac{C^2+D^2-4AF}{4A^2}$, thus, with $C^2+D^2>4AF$ the equation represent a circle centered at $(-\frac{C}{2A}, -\frac{D}{2A})$, with radius $\frac{\sqrt{C^2+D^2-4AF}}{2A}$ ### Q4 ### By completing squares, we should get $3(x-1)^2 + 2(y+2)^2=36$, Divide by 36, we have $\frac{(x-1)^2}{12} + \frac{(y+2)^2}{18} = 1$ So $(p, q) = (1,-2)$, $a = 2\sqrt{3}$, $b=3\sqrt{2}$, $c=\pm \sqrt{6}$ However, this is a vertical major axis ellipse, so, - Center: $(1,-2)$ - Foci: $(1, -2-\sqrt{6})$ and $(1, -2+\sqrt{6})$ - Eccentricity: $\frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2}$ ### Q5 ### 1. As $(a,0)$ is one point of the ellipse, we have $sum\ of\ distance\ to\ foci= a-c + (a+c) = 2a$ 2. Also we have $(0, b)$ is one point of ellipse on y-axis, with Pythagorean theorem, we have $b^2 + c^2 = a^2$, where a refers to distance from point $(0,d)$ to $(c, 0)$, which is half of $2a$ due to symmetry. So we have $c^2 = a^2 - b^2$ 3. Thus based on definition, we have $$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2a$$ $\implies \sqrt{(x+c)^2+y^2} = 2a - \sqrt{(x-c)^2+y^2}$ $\implies x^2+2cx+c^2+y^2=4a^2+x^2-2cx+c^2+y^2-4a\sqrt{(x-c)^2+y^2}$ $\implies \sqrt{(x-c)^2+y^2} = a - \frac{cx}{a}$ $\implies x^2 - 2cx + c^2 + y^2 = a^2 - 2cx + \frac{c^2}{a^2}x^2$ $\implies a^2 - b^2 + y^2 = a^2 - \frac{b^2}{a^2}x^2$ $\implies \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, as required. #### Q6 Omitted, just plug-in z = 2, and do some algebra.