Chapter 11.5, Problem 2E

(a)

To determine

To fill: The statement “Consider the equation $A{x}^2+Bxy+c{y}^2+Dx+Ey+D=0$ In general, the graph of this equation is a             .”

Answer to Problem 2E

Consider the equation $A{x}^2+Bxy+c{y}^2+Dx+Ey+D=0$. In general, the graph of this equation is a conic.

Explanation of Solution

Definition used:

Definition 1:

“The general equation of conic is of the form, $A{x}^2+Bxy+c{y}^2+Dx+Ey+D=0$ where $A,B,C,D,E$ and $F$ are constants.”

Calculation:

By the definition stated above, it is clear that the equation $A{x}^2+Bxy+c{y}^2+Dx+Ey+D=0$ represents a conic in general.

(b)

To determine

To fill: The statement “To eliminate the xy-term from the equation $A{x}^2+Bxy+c{y}^2+Dx+Ey+D=0$, we rotate the axes through an angle $\varphi$ that satisfies    $\cos 2\varphi =$ .”

Answer to Problem 2E

To eliminate the xy-term from the equation $A{x}^2+Bxy+c{y}^2+Dx+Ey+D=0$, we rotate the axes through an angle $\varphi$ that satisfies $\cot 2\varphi =\frac{A-C}{B}$.

Explanation of Solution

Definition used:

Definition 2:

“Rotation of co-ordinate 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 $A{x}^2+Bxy+c{y}^2+Dx+Ey+F=0$”.

Calculation:

The general equation of conic is $A{x}^2+Bxy+c{y}^2+Dx+Ey+D=0$ (1)

Now, the rotation of axes through an acute angle $\varphi$ makes the axes on to its new position X and Y. therefore, x and y in terms of X and Y given as follows.

$x=X\cos \varphi -Y\sin \varphi$ (2)

$y=X\sin \varphi +Y\cos \varphi$ (3)

Substitute the equations (1) and (2) in equation (1),

$\left\{\begin{array}{l}A{\left(X\cos \varphi -y\sin \varphi \right)}^2+B\left(X\cos \varphi -y\sin \varphi \right)\left(X\sin \varphi +y\cos \varphi \right)\\ +C{\left(X\sin \varphi +y\cos \varphi \right)}^2+D\left(X\cos \varphi -y\sin \varphi \right)+E\left(X\sin \varphi +y\cos \varphi \right)+F\end{array}\right\}=0$

$\left\{\begin{array}{l}A\left(X^2{\cos }^2\varphi -2XY\cos \varphi \sin \varphi +Y^2{\sin }^2\varphi \right)+\\ B\left(X^2\cos \varphi \sin \varphi +XY{\cos }^2\varphi -XY{\sin }^2\varphi -Y^2\sin \varphi \cos \varphi \right)\\ +C\left(X^2{\cos }^2\varphi +2XY\cos \varphi +Y^{2}co{s}^2\varphi \right)+D\left(X\cos \varphi -Y\sin \varphi \right)+\\ E\left(X\sin \varphi +y\cos \varphi \right)+F\end{array}\right\}=0$ (4)

Consider the coefficient of $x^2$, $xy$, $y^2$, $x$, $y$ and constant terms of above equation and assume $A^{\prime }$, $B^{\prime }$, $C^{\prime }$, $D^{\prime }$, $E^{\prime }$ and $F^{\prime }$ respectively their values are as follows.

$A^{\prime }=A{\cos }^2\varphi +B\cos \varphi \sin \varphi +C{\sin }^2\varphi$ (5)

$B^{\prime }=2\left(C-A\right)\sin \varphi \cos \varphi +B\left({\cos }^2\varphi -{\sin }^2\varphi \right)$ (6)

$C^{\prime }=A{\sin }^2\varphi -B\sin \varphi \cos \varphi +C{\cos }^2\varphi$ (7)

$D^{\prime }=D\cos \varphi +E\sin \varphi$ (8)

$E^{\prime }=-D\sin \varphi +E\cos \varphi$ (9)

$F^{\prime }=F$ (10)

Now, the equation (4) becomes,

$A^{\prime }{x}^2+B^{\prime }xy+c^{\prime }{y}^2+D^{\prime }x+E^{\prime }y+F^{\prime }=0$ (11)

From equation (1), it is clear that if $B^{\prime }=0$, then $xy$ term will be eliminated.

Therefore, $B^{\prime }=0$ in the equation (6), $2\left(C-A\right)\sin \varphi \cos \varphi +B\left({\cos }^2\varphi -{\sin }^2\varphi \right)=0$.

Substitute $2\sin \varphi \cos \varphi =\sin 2\varphi$, ${\cos }^2\varphi -{\sin }^2\varphi =\cos 2\varphi$ in above equation.

$\begin{array}{l}\left(C-A\right)\sin 2\varphi +B\cos 2\varphi =0\\ B\cos 2\varphi =-\left(C-A\right)\sin 2\varphi \\ B\cos 2\varphi =\left(A-C\right)\sin 2\varphi \\ \frac{\cos 2\varphi }{\sin 2\varphi }=\frac{A-C}{B}\end{array}$

Simplify further,

$\cot 2\varphi =\frac{A-C}{B}$.

Thus, to eliminate the xy-term from the equation $A{x}^2+Bxy+c{y}^2+Dx+Ey+D=0$, rotate the axes through an angle $\varphi$ that satisfies, $\cot 2\varphi =\frac{A-C}{B}$.

(c)

To determine

To fill: The statement “The discriminant of this equation is         .

If the discriminant is zero, the graph is a                 .

If it is negative, the graph is an               .

If it is positive, the graph is a                  .”

Answer to Problem 2E

The discriminant of this equation is $B^2-4AC$.

If the discriminant is zero, the graph is a parabola.

If it is negative, the graph is an ellipse.

If it is positive, the graph is a hyperbola.

Explanation of Solution

Definition used:

Definition 2:

“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, $B^2-4AC>0$.”

Calculation:

The general equation of conic is $A{x}^2+Bxy+c{y}^2+Dx+Ey+F=0$.

The discriminant of the above equation is $B^2-4AC$.

By the definition 2 stated above the graph of the equation (1) is a parabola if $B^2-4AC=0$, is an ellipse if $B^2-4AC<0$ and a hyperbola if $B^2-4AC>0$.

That is, If the discriminant is zero, then graph is a parabola.

If it is negative, the graph is an ellipse.

If it is positive, the graph is a hyperbola.