Chapter 11, Problem 13T

1. (a) Use the discriminant to determine whether the graph of the following equation is a parabola, an ellipse, or a hyperbola:

$5x^2+4xy+2y^2=18$

1. (b) Use rotation of axes to eliminate the xy-term in the equation.
2. (c) Sketch a graph of the equation.
3. (d) Find the coordinates of the vertices of this conic (in the xy-coordinate system).

To determine

Whether the graph of the equation $5x^2+4xy+2y^2=18$ is a parabola, an ellipse or a hyperbola by discriminant.

Answer to Problem 13T

The graph of the equation $5x^2+4xy+2y^2=18$ is an ellipse.

Explanation of Solution

Definition used:

Definition 1:

“The graph of the equation $A{x}^2+Bxy+c{y}^2+Dx+Ey+F=0$ is

1. (i) A parabola if the discriminant, $B^2-4AC=0$
2. (ii) An ellipse if the discriminant,  $B^2-4AC<0$
3. (iii) A hyperbola if the discriminant, $B^2-4AC>0$”.

Given the equation of conic is $5x^2+4xy+2y^2=18$ (1)

Compare the equation (1) with the equation of conic $A{x}^2+Bxy+c{y}^2+Dx+Ey+F=0$,

it is clear that, $A=5$, $B=4$, $C=2$.

Now, the discriminant,

$B^2-4AC={\left(4\right)}^2-4\left(4\right)\left(2\right)$

$B^2-4AC=16-40$

$B^2-4AC=-24<0$

Since $B^2-4AC<0$ by the definition stated above, equation (1) represents a ellipse

Therefore, the equation $5x^2+4xy+2y^2=18$ is an ellipse.

To determine

To eliminate: The xy-term is eliminated from of conic $5x^2+4xy+2y^2=18$.

Answer to Problem 13T

The xy-term is eliminated from of conic $5x^2+4xy+2y^2=18$, and the eliminated equation is $42X^2+106Y^2-16X=0$.

Explanation of Solution

Definition used:

Definition 2:

“Rotation of coordinate axes through an acute angle $\varphi$ with the condition that, $\cot 2\varphi =\frac{A-C}{B}$, eliminates the xy-term from the general equation of conic $A{x}^2+Bxy+c{y}^2+Dx+Ey+F=0$”.

Definition 3:

“Suppose x and y axes in a coordinate plane are rotated through an acute angle $\varphi$ to produce the X and Y axes. Then the coordinates of a point $\left(x,y\right)$ in xy-plane in terms of X and Y are given by,

$x=X\cos \varphi -Y\sin \varphi$

$y=X\sin \varphi +y\cos \varphi$”.

Given the equation of conic is $5x^2+4xy+2y^2=18$ (1)

Compare the equation (1) with the general equation of conic $A{x}^2+Bxy+c{y}^2+Dx+Ey+F=0$,

It is clear that, $A=5$, $B=4$, $C=2$.

By the definition stated above, the rotation of coordinate axes through an angle $\varphi$ with $\cot 2\varphi =\frac{A-C}{B}$ eliminates xy-term from the equation (1).

Therefore,

$\cot 2\varphi =\frac{5-2}{4}$

$\cot 2\varphi =\frac{3}{4}$

That is, opposite side = 4

Adjacent side 3 as shown in the figure 1.

By Pythagoras theorem,

$AC=\sqrt{{(AB)}^2+{\left(BC\right)}^2}$

$AC=\sqrt{3^2+4^2}$

$AC=\sqrt{9+16}$

$AC=\sqrt{25}$

$AC=5$.

Therefore, $\cos 2\varphi =\frac{AB}{AC}$

$\cos 2\varphi =\frac{3}{5}$ $2\varphi$

Use half angle formulas,

$\cos \varphi =\sqrt{\frac{1+\cos 2\varphi }{2}}$

$\cos \varphi =\sqrt{\frac{1+\frac{3}{5}}{2}}$

$\cos \varphi =\sqrt{\frac{8}{10}}$

$\cos \varphi =\sqrt{\frac{4}{5}}$

$\cos \varphi =\frac{2}{\sqrt{5}}$.

Similarly,

$\sin \varphi =\sqrt{\frac{1-\cos 2\varphi }{2}}$

$\sin \varphi =\sqrt{\frac{1-\frac{3}{5}}{2}}$

$\sin \varphi =\sqrt{\frac{2}{10}}$

$\sin \varphi =\frac{1}{\sqrt{5}}$

Therefore, rotation of axes by $\sin \varphi =\frac{1}{\sqrt{5}}$ eliminates xy-term from equation (1).

From the definition 3 stated above, if the xy-coordinates of equation (1) rotated through $\sin \varphi =\frac{1}{\sqrt{5}}$ then, xy- coordinates in terms of X and Y are,

$x=X\cos \varphi -Y\sin \varphi$

$y=X\sin \varphi +Y\cos \varphi$ with $\cos \varphi =\frac{2}{\sqrt{5}}$, $\sin \varphi =\frac{1}{\sqrt{5}}$.

$x=\frac{2}{\sqrt{5}}X-\frac{1}{\sqrt{5}}Y$

$x=\frac{1}{\sqrt{5}}\left(2X-Y\right)$ (2)

And,

$y=\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y$

$y=\frac{1}{\sqrt{5}}\left(X+2Y\right)$ (3)

Substitute equation (2) and (3) in equation (1), then obtain the resulting equation,

$5x^2+4xy+2y^2=18$

$5\times \frac{1}{5}{\left(2X-Y\right)}^2+4\times \frac{1}{5}\left(2X-Y\right)\left(X+2Y\right)+2\times \frac{1}{5}{\left(X+2Y\right)}^2=18$

Multiply both sides by $5$,

$5\times {\left(2X-Y\right)}^2+4\times \left(2X-Y\right)\left(X+2Y\right)+2\times {\left(X+2Y\right)}^2=90$

$5\left(8X^2-4XY+Y^2\right)+4\left(2X^2+4XY-XY-2Y^2\right)+2\left(X^2+4XY+4Y^2\right)=90$

$40X^2-20XY+5Y^2+14X^2+12XY-8Y^2+2X^2+8XY+8Y^2=90$

$56X^2+5Y^2=90$. Which is an ellipse.

Therefore, the equation of conic after the elimination of xy-term is

$56X^2+5Y^2=90$.

To determine

To sketch:

The graph of $9x^2+24xy+16y^2=25$,

Answer to Problem 13T

The graph of the equation $9x^2+24xy+16y^2=25$ is as shown below in figure1.

Explanation of Solution

The graph of the equation $5x^2+4xy+2y^2=18$ is drawn by the use of graphing device and represented as shown below in figure 1.

From the graph, it is clear that the graph of the equation $5x^2+4xy+2y^2=18$ is an ellipse with transverse axis as the line makes an angle $\sin \varphi =\frac{1}{\sqrt{5}}$ with x-axis. Thus, $5x^2+4xy+2y^2=18$ represent a rotated ellipse.

To determine

To find: The find the vertices of the conic $5x^2+4xy+2y^2=18$ in xy coordinates.

Answer to Problem 13T

The vertices of the conic $5x^2+4xy+2y^2=18$ in terms of xy coordinates are $V_1\left(3\sqrt{\frac{2}{5}},-6\sqrt{\frac{2}{5}}\right)$ and $V_2\left(-3\sqrt{\frac{2}{5}},6\sqrt{\frac{2}{5}}\right)$.

Explanation of Solution

Definition used:

Definition 4:

“The equation of the ellipse with center  at the origin, vertices $V(0,\pm a)$ and foci $F(0,\pm c)$ is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, where $c=\sqrt{a^2-b^2},a>b>0$”.

The equation of conic $5x^2+4xy+2y^2=18$ after the elimination of xy-term is

$56X^2+5Y^2=90$ (4)

Now, the equation (4) can be written by dividing both sides by 90 as follows,

$\frac{56}{90}{X}^2+\frac{5}{90}{Y}^2=1$

$\frac{7}{15}{X}^2+\frac{1}{18}{Y}^2=1$

$\frac{X^2}{\frac{15}{7}}+\frac{Y^2}{18}=1$ (5)

The equation (5) is of the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$.

Therefore, compare $\frac{X^2}{\frac{15}{7}}+\frac{Y^2}{18}=1$ with $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ and conclude,

$a^2=18$ , $b^2=\frac{15}{7}$.

Take positive square root for both the equations,

$a=\sqrt{18}$ , $b=\sqrt{\frac{15}{7}}$.

$a=\sqrt{9\times 2}$, $b=\sqrt{\frac{15}{7}}$.

$a=3\sqrt{2}$, $b=\sqrt{\frac{15}{7}}$.

Clearly, $a=5$, $b=4$ and $a>b>0$. Hence, by the above definition 4, the vertices of the ellipse are $\frac{X^2}{\frac{15}{7}}+\frac{Y^2}{18}=1$ $V_1(0,-3\sqrt{2})$ and $V_2(0,3\sqrt{2})$.

Therefore, the vertices of the conic $5x^2+4xy+2y^2=18$ in terms of X and Y are $V_1(0,-3\sqrt{2})$ and $V_2(0,3\sqrt{2})$.

From the equations (2) and (3) the x and y coordinates in terms of X and Y are,

$x=\frac{1}{\sqrt{5}}\left(2X-Y\right)$

$y=\frac{1}{\sqrt{5}}\left(X+2Y\right)$

Apply the vertices of the conic $5x^2+4xy+2y^2=18$ in terms of X and Y on the above coordinates,

$V_1\left(\frac{1}{\sqrt{5}}\left(2(0)-(-3\sqrt{2})\right),\frac{1}{\sqrt{5}}\left(0+2(-3\sqrt{2})\right)\right)$ (Since $X=0$, $Y=-3\sqrt{2}$ for $V_1$)

$V_1\left(\frac{3\sqrt{2}}{\sqrt{5}},\frac{-6\sqrt{2}}{\sqrt{5}}\right)$.

$V_1\left(3\sqrt{\frac{2}{5}},-6\sqrt{\frac{2}{5}}\right)$.

And,

$V_2\left(\frac{1}{\sqrt{5}}\left(2(0)-(3\sqrt{2})\right),\frac{1}{\sqrt{5}}\left(0+2(3\sqrt{2})\right)\right)$ (Since $X=0$, $Y=3\sqrt{2}$ for $V_2$)

$V_2\left(\frac{-3\sqrt{2}}{\sqrt{5}},\frac{6\sqrt{2}}{\sqrt{5}}\right)$.

$V_2\left(-3\sqrt{\frac{2}{5}},6\sqrt{\frac{2}{5}}\right)$.

Thus, the vertices of the conic $5x^2+4xy+2y^2=18$ in terms of xy coordinates are $V_1\left(3\sqrt{\frac{2}{5}},-6\sqrt{\frac{2}{5}}\right)$ and $V_2\left(-3\sqrt{\frac{2}{5}},6\sqrt{\frac{2}{5}}\right)$.