### Author Topic: how to solve question 19 in section 1.2 in the textbook  (Read 574 times)

#### caowenqi

• Newbie
• Posts: 4
• Karma: 0
##### how to solve question 19 in section 1.2 in the textbook
« on: September 23, 2020, 04:43:35 PM »
how to solve question 19 in section 1.2 in the textbook
I didn't understand the solution for 19(a) and (c). Thanks a lot!

#### Darren Zhang

• Full Member
• Posts: 22
• Karma: 13
##### Re: how to solve question 19 in section 1.2 in the textbook
« Reply #1 on: September 26, 2020, 08:06:10 PM »
Expanding the restriction reveals.

$\lvert z - p\rvert = cx \Rightarrow \sqrt{(x-p)^2 + y ^ 2} = cx \Rightarrow x^2 - 2xp + p^2 + y^2 = c^2x^2, x > 0$

$\Rightarrow (1 - c^2)x^2 - 2xp + y^2 = 0$

We know the sign of the quadratic component is what determines the behaviour. Thus

$c \in (0,1) \Rightarrow (1 − c_{2}) > 0 \Rightarrow \frac{{x'}^2}{a^2} + y^2 = 1 \Rightarrow$ ellipse
$c = 0 \Rightarrow (1 − c_{2}) = 0 \Rightarrow x = \frac{y^2}{2p} \Rightarrow$ parabola
$c \in (1,\infty) \Rightarrow (1 − c_{2}) < 0 \Rightarrow -\frac{{x'}^2}{a^2} + y^2 = 1 \Rightarrow$ hyperbola

where x' is the appropriate translation of x.

#### RunboZhang

• Sr. Member
• Posts: 51
• Karma: 0
##### Re: how to solve question 19 in section 1.2 in the textbook
« Reply #2 on: September 27, 2020, 09:42:31 AM »
Hello Darren, I have two questions regarding your solution. Firstly, why did you exclude p^2 in equation (1-c^2)x^2-2xp+y^2=0 where p is a real number? And secondly, when you got the solution for eclipse and hyperbola, how did you get 1 on RHS? I thought RHS would be an expression with respect to x and c and it is not necessary to be 1 right?

#### Victor Ivrii

• Elder Member
• Posts: 2569
• Karma: 0
##### Re: how to solve question 19 in section 1.2 in the textbook
« Reply #3 on: September 27, 2020, 11:37:07 AM »
Darren, you may want to fix: replace $c_2$ by $c^2$.

#### Victor Ivrii

Darren, you may want to fix: replace $c_2$ by $c^2$.