---
# System prepended metadata

title: Calculas Scholarship T2W1
tags: [Calculus Scholarship]

---

---

title: Calculas Scholarship T2W1 
tags: Calculus Scholarship 
description: Term 2 Week 1 
type: pdf
slideOptions:
 theme: white

---

###### tags: `Calculus Scholarship`

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### 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

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### 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.


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### 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 $\frac{\sqrt{C^2+D^2-4AF}}{2A}$


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### 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: $\frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2}$

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### 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.

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#### Q6 Omitted, just plug-in z = 2, and do some algebra.






