---

title: Calculas Scholarship T2W2
tags: Calculus Scholarship
description: Term 2 Week 2
type: pdf
slideOptions:
theme: white

---

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

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### Q1 ###

For the first subquestion, plug-in the line equation to the conic, we have $(8x+1)^2=32x$, solve this quadratic get the x-coordinate of the tangent point. Using x-coordinate in line equation to get y-coordinate of the point;

The second subquestion, $25x^2-9(mx+3)^2=225$, so we have,
$$25x^2-9m^2x^2-54mx-81-225=0$$, so $a = 25-9m^2$, $b=-54m$, $c=306$. Therefore, "tangent" means $b^2-4ac=0\implies54^2m^2+36\cdot6\cdot51m^2=4\cdot25\cdot6\cdot51$
Solve quadratic of m, get two values of m. Plug-in each m into $25x^2-9(mx+3)^2=225$, get one x-coordinate of tangential point.

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### Q2 ###

1). For parabola, $(mx+c)^2 = 4ax\implies m^2x^2+2mcx+c^2=4cmx$
$\implies m^2x^2-2mcx+c^2=0\implies (mx-c)^2=0$, which have one root $\frac{c}{m}$, or $\frac{a}{m^2}$. so it tangent the conic at $(\frac{a}{m^2},\frac{2a}{m})$

I will just do the ellipse, the rest follows the same method.

3). with $\frac{x^2}{a^2}+\frac{(mx+c)^2}{b^2}=1\implies b^2x^2+a^2(m^2x^2+2mcx+a^2m^2+b^2)=a^2b^2$
$\implies(a^2m^2+b^2)x^2+2mca^2x+a^4m^2=0$
$\implies c^2x^2+2mca^2x+a^4m^2=0\implies(cx+a^2m)^2=0$
which have only one root $-\frac{a^2m}{c}$.

solving $y = m\cdot-\frac{a^2m}{c}+c =-\frac{m^2a^2}{c}+c=-\frac{(c^2-b^2)a^2}{a^2c}+c=\frac{b^2-c^2}{c}+c=\frac{b^2}{c}$

So this proved that the line is tangent to the ellipse at $(-\frac{a^2m}{c}, \frac{b^2}{c})$

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### Q3 ###

Applie PSR to all conics:

1). $x^2+y^2=5 \implies (x-0)^2+(y-0)^2=5$, so apply PSR at $(1,2)$, we have $(x_1-0)(x-0)+(y_1-0)(y-0)=5$
$\implies(1-0)x+(2-0)y=5\implies x+2y=5$

2).$3x^2-y^2-2=0\implies \frac{(x-0)^2}{\frac{2}{3}}-\frac{(y-0)^2}{2} = 1$, so applie PSR at $(3,-5)$, we have $\frac{(x_1-0)(x-0)}{\frac{2}{3}}-\frac{(y_1-0)(y-0)}{2} = 1$
$\implies\frac{(3-0)(x-0)}{\frac{2}{3}}-\frac{(-5-0)(y-0)}{2} = 1$
$\implies9x+5y-2=0$ <hint, you can also apply PSR without transform into standard equaiton.>

3).$4x^2-3y^2+2x-y-26=0$, this time, we apply PSR directly rather than transform to standard equation. At $(-3,1)$, we have
$4(x_1x)-3(y_1y)+2(\frac{x+x_1}{2})-(\frac{y+y_1}{2})-26=0$
$\implies-12x-3y+x-3-\frac{y}{2}-\frac{1}{2}-26=0$
$\implies-22x-7y-59=0$

I hope you have learnt the PSR. Use the PSR to do the next two questions, and chech your correctness with Desmos

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### Q4 ###

1. Using PSR, get tangent at $(10, 10)$, which is $\frac{7(x-3)}{a^2}+\frac{12(y+2)}{b^2}=1$
2. Put it into Point-slope form:
$\implies7b^2x-21b^2+12a^2(y+2)=a^2b^2$
$\implies12a^2(y+2)=-7b^2x+21b^2+a^2b^2$
$\implies12a^2(y+2)=-7b^2(x-3-\frac{a^2}{7})$
$\implies(y+2)=-\frac{7b^2}{12a^2}(x-(3+\frac{a^2}{7}))$
3. Get the normal $(y+2)=\frac{12a^2}{7b^2}(x-(3+\frac{a^2}{7}))$
4. Then solve for a and b using points (0,0) and (10, 10), omitted.

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### Q5 ###

One of the boring standard solution:


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### Q6 ###

The first circle can be $(x-8)^2+(y-10)^2=7^2$,
The second circle can be $(x+4)^2+(y-5)^2=6^2$

The distance between two centers is $\sqrt{12^2+5^2}=13$, just the sum of radius of two circle, as required.