---
# System prepended metadata

title: Calculas Scholarship T2W3
tags: [Calculus Scholarship]

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title: Calculas Scholarship T2W3 
tags: Calculus Scholarship 
description: Term 2 Week 3 
type: pdf
slideOptions:
 theme: white

---

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

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### Q1 ###

Using trigonometry identity $sec^2(\theta)-tan^2(\theta) = 1$, we have 
$(x+1)^2- (\frac{y-2}{2})^2 = 1$
clearly a hypobola centered at $(-1,2)$, with foci $(-\sqrt{5}-1,2)$ and $(\sqrt{5}-1,2)$

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### Q2 ###

This time we will use Andrew's favorate method for the first question. Because it will be faster :)


a). Apply differential operator on both sides, we have $\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x^2}{a^2}+\frac{y^2}{b^2})=\frac{\mathrm{d}}{\mathrm{d}x}1$
$\implies \frac{2x}{a^2} + \frac{2y}{b^2}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}=0$
$\implies\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{b^2x}{a^2y}$, so at the point (m,n), the slope of the tangent would be $-\frac{b^2m}{a^2n}$.

b).Another way of getting the slope of the tangent is from two points given, and it must equals to the slope we get from a). So we have 
$\frac{n}{m-c} = \frac{-b^2m}{a^2n} \implies a^2n^2+b^2m^2=b^2cm\implies\frac{n^2}{b^2}+\frac{m^2}{a^2}=\frac{mc}{a^2}$
As $(m,n)$ must lies on the ellipse, so we have $\frac{mc}{a^2} = 1\implies m=\frac{a^2}{c}. \blacksquare$

c). Plug-in the $m=\frac{a^2}{c}$ into the ellipes equation, solve for n, omitted.

d). As $c \rightarrow a$, from b), $m \rightarrow a$, from c), $n\rightarrow0$.

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### Q3 ###

(i). The itention of this question is trying to emphasize on the difference between the parameter in the Parametric Equation and the actrual angle from the ellipse to the center.

A posible way to help you understand is to replace $\theta$ to $t$, so we have $tan(\phi) = \frac{bsin(t)}{acos(t)}=\frac{b}{a}tan(t)$

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### Q4 ###

With Parametric Equation to the parabola, we have $x(t) = at^2$ and $y(t)=2at$, so
$\frac{\mathrm{d}x}{\mathrm{d}t} = 2at$ and $\frac{\mathrm{d}y}{\mathrm{d}t} = 2a$, thus
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t}\cdot \frac{\mathrm{d}t}{\mathrm{d}x}  =2a\cdot\frac{1}{2at} = \frac{1}{t} = tan(\angle ACF)$

by double angle formular
$tan(2\angle ACF) = \frac{\frac{2}{t}}{1-\frac{1}{t^2}}=\frac{2t}{t^2-1}$

For the angle AFx, we have $tan(\angle AFx) = \frac{2at}{at^2-a}=\frac{2t}{t^2-1}$

This shows that $\angle AFx = 2\angle ACF\implies\angle BFA = 2\angle ACF$ by alternative angles.

- Another way using Geometric method would be told in the workshop


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### Q5 ###

Let the equation $y = mx + \Bbb{C}$ represent the collection of parallal chords. Be aware that as this is a set of parallal chords, their slope must be identical, with different y-intercept, i.e. $\Bbb{C}$ represent a variable whereas $m$ represent a fixed constant.

now we can use the set of equations $y = mx + \Bbb{C}$ to subtitude the ellipse equation.

$\frac{x^2}{a^2}+\frac{(mx+\Bbb{C})^2}{b^2}=1$
$\implies b^2x^2 + a^2m^2x^2+a^2\Bbb{C}^2+2a^2m\Bbb{C}x=a^2b^2$
$\implies(b^2+a^2m^2)x^2+ 2a^2m\Bbb{C}x + a^2(\Bbb{C}^2-b^2)=0$

The roots of the quadratic above with different $c\in\Bbb{C}$ **represent the  x-coordinate of the chords**.

Let the set $(\Bbb{P},\Bbb{Q})$ be the coordinates of midpoints of chords. So $\Bbb{P} = \frac{X_1 + X_2}{2}$, where $X_1, X_2$ are two roots from each of the quadratic equation in the set of equations above. Thus by **Sum of roots** theorem, we have $\Bbb{P}=\frac{-a^2\Bbb{c}m}{b^2+a^2m^2}\implies\Bbb{C}=\frac{b^2+a^2m^2}{-a^2m}\Bbb{P}$
As $(\Bbb{P},\Bbb{Q})$ must satisfy $y = mx + \Bbb{C}$, so we have 
$\Bbb{Q}=m\Bbb{P}-\frac{b^2+a^2m^2}{a^2m}\Bbb{P}\implies a^2m\Bbb{Q}=a^2m^2\Bbb{P}-b^2\Bbb{P}-a^2m^2\Bbb{P}=-b^2\Bbb{P}$
or $a^2m\Bbb{Q}+b^2\Bbb{P}=0$ From this linear relation we can conclude that the midpoints of chords must pass through the origin.
