how to solve question 19 in section 1.2 in the textbook

[caowenqi]

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]

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]

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]

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

[Victor Ivrii]

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


Runbo, what  about division?