Question

the equation of an ellipse with center shifted to (h,k) is \\(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\\) (12.32). i) (81 + 16x^{2}-128x + 25y^{2}+150y = 0) ii) (9x^{2}+4y^{2}-8y = 32) iii) (x^{2}+y^{2}-2y = 2) iv) (8x^{2}+8x + 5y^{2}-20y = 18) f) for the cases in part e), determine the lengths of the semi - major and semi - minor axes (or the radius if circular). g) in the equation: (x^{2}(1 - e^{2})+y^{2}+a^{2}(e^{2}-1)=0) (12.33) the eccentricity is still given by (e = c/a), but now it can be either greater than or less than one. for (e < 1), show that eq. (12.33) gives the standard equation for an ellipse, where (a), (b), and (c) satisfy the equation you found in part b). for (e>1), show that it instead gives (\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1) (12.34) where the alternative relation (b^{2}=c^{2}-a^{2}) now applies. equation (12.34) describes a hyperbola, defined as the set of points where the absolute value of the difference of the distances from two points (the focal points) is a constant.

Explanation:

Step1: Rewrite the general - form equations to standard form

For an ellipse, the standard form is $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$ (for non - circular case) and for a circle, it is $(x - h)^2+(y - k)^2=r^2$.

Case (i):

Given $81 + 16x^{2}-128x + 25y^{2}+150y = 0$.
First, group the $x$ and $y$ terms: $16(x^{2}-8x)+25(y^{2}+6y)=- 81$.
Complete the square for $x$ and $y$:
$16(x^{2}-8x + 16)-16\times16+25(y^{2}+6y + 9)-25\times9=-81$.
$16(x - 4)^{2}+25(y + 3)^{2}=256 + 225-81$.
$16(x - 4)^{2}+25(y + 3)^{2}=399$.
$\frac{(x - 4)^{2}}{\frac{399}{16}}+\frac{(y + 3)^{2}}{\frac{399}{25}}=1$.
The semi - major axis $a=\sqrt{\frac{399}{16}}\approx4.99$ and semi - minor axis $b=\sqrt{\frac{399}{25}}\approx3.99$.

Case (ii):

Given $9x^{2}+4y^{2}-8y = 32$.
Group the $y$ terms: $9x^{2}+4(y^{2}-2y)=32$.
Complete the square for $y$: $9x^{2}+4(y^{2}-2y + 1)-4=32$.
$9x^{2}+4(y - 1)^{2}=36$.
$\frac{x^{2}}{4}+\frac{(y - 1)^{2}}{9}=1$.
The semi - major axis $a = 3$ and semi - minor axis $b = 2$.

Case (iii):

Given $x^{2}+y^{2}-2y=2$.
Complete the square for $y$: $x^{2}+(y^{2}-2y + 1)-1=2$.
$x^{2}+(y - 1)^{2}=3$.
This is a circle with radius $r=\sqrt{3}\approx1.73$.

Case (iv):

Given $8x^{2}+8x + 5y^{2}-20y=18$.
Group the $x$ and $y$ terms: $8(x^{2}+x)+5(y^{2}-4y)=18$.
Complete the square for $x$ and $y$:
$8(x^{2}+x+\frac{1}{4})-8\times\frac{1}{4}+5(y^{2}-4y + 4)-5\times4=18$.
$8(x+\frac{1}{2})^{2}+5(y - 2)^{2}=18 + 2+20$.
$8(x+\frac{1}{2})^{2}+5(y - 2)^{2}=40$.
$\frac{(x+\frac{1}{2})^{2}}{5}+\frac{(y - 2)^{2}}{8}=1$.
The semi - major axis $a=\sqrt{8}\approx2.83$ and semi - minor axis $b=\sqrt{5}\approx2.24$.

Step2: Analyze the equation $x^{2}(1 - e^{2})+y^{2}+a^{2}(e^{2}-1)=0$

When $e\lt1$:

Let $e=\frac{c}{a}$ where $c\lt a$.
Rewrite the equation $x^{2}(1 - e^{2})+y^{2}+a^{2}(e^{2}-1)=0$ as $x^{2}(1-\frac{c^{2}}{a^{2}})+y^{2}-a^{2}(1 - \frac{c^{2}}{a^{2}})=0$.
$\frac{a^{2}-c^{2}}{a^{2}}x^{2}+y^{2}-(a^{2}-c^{2})=0$.
Since $b^{2}=a^{2}-c^{2}$, we have $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, which is the standard equation of an ellipse.

When $e\gt1$:

Rewrite the equation $x^{2}(1 - e^{2})+y^{2}+a^{2}(e^{2}-1)=0$ as $y^{2}-(e^{2}-1)x^{2}=a^{2}(e^{2}-1)$.
Let $b^{2}=c^{2}-a^{2}$ (where $c\gt a$ since $e=\frac{c}{a}\gt1$).
The equation becomes $\frac{y^{2}}{a^{2}(e^{2}-1)}-\frac{x^{2}}{a^{2}} = 1$ or $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$ (after substituting $b^{2}=c^{2}-a^{2}$), which is the standard equation of a hyperbola.

Answer:

For part (f):

• (i) Semi - major axis $\approx4.99$, semi - minor axis $\approx3.99$.
• (ii) Semi - major axis $a = 3$, semi - minor axis $b = 2$.
• (iii) Radius $r=\sqrt{3}\approx1.73$.
• (iv) Semi - major axis $\approx2.83$, semi - minor axis $\approx2.24$.

For part (g): When $e\lt1$, the equation $x^{2}(1 - e^{2})+y^{2}+a^{2}(e^{2}-1)=0$ gives the standard equation of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$; when $e\gt1$, it gives the standard equation of a hyperbola $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$.