Chapter A5.3, Problem 37E

(a)

To determine

To find: the conic section has the given properties.

Answer to Problem 37E

Here, the given conic section symmetric to the origin as $A{x}^2+Bxy+C{y}^2+F=0$ .

Explanation of Solution

Given: $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

Symmetric with respect to the origin.

Calculation:

Symmetry about the origin means that $\left(-x,-y\right)$ lies on the graph whenever $\left(x,y\right)$ does.

Therefore, if

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Then,

  $A{\left(-x\right)}^2+B\left(-x\right)\left(-y\right)+C{\left(-y\right)}^2+D\left(-x\right)+Ey+F=0$

Adding the above two equations and then dividing by $2$ ,

  $A{x}^2+Bxy+C{y}^2+F=0$

Conclusion:

Hence, the conic section symmetric to the origin is as $A{x}^2+Bxy+C{y}^2+F=0$ .

(b)

To determine

To find: the conic section has the given properties.

Answer to Problem 37E

Here, the given conic section passes the given point by $-F{x}^2+Bxy+C{y}^2+F=0$ .

Explanation of Solution

Given; Passes through the point $\left(1,0\right)$ .

Calculation:

Since the point $\left(1,0\right)$ lies on the curve, $\left(x=1\right)$ and $\left(y=0\right)$ should satisfy the above equation. Therefore,

  $\begin{array}{l}A{\left(1\right)}^2+B\left(1\right)\left(0\right)+C{\left(0\right)}^2+F=0\\ \Rightarrow A+F=0\\ \Rightarrow A=-F\end{array}$

Substituting the value of $A=-F$ , the equation is reduced to

  $-F{x}^2+Bxy+C{y}^2+F=0$ .

Conclusion:

Hence, the given conic section passes the given point by $-F{x}^2+Bxy+C{y}^2+F=0$ .

(c)

To determine

To solve: the given conic section is tangent.

Answer to Problem 37E

Here, the given conic section is tangent by the curve $x^2+4xy+5y^2-1=0$ .

Explanation of Solution

Given; Tangent tot the line $y=1$ to the point $(-2,1)$ .

Calculation:

By implicit differentiation with respect to $x$ , the equation becomes

  $-2Fx+By+Bxy\text{'}+2Cyy\text{'}=0$

For the line $y=1$

  $y\text{'}=0$

Now, being given that at $\left(x,y\right)=\left(-2,1\right),y\text{'}=0$ ,

  $\begin{array}{l}-2F\left(-2\right)+B\left(1\right)+B\left(-2\right)\left(0\right)+2C\left(1\right)\left(0\right)=0\\ \Rightarrow 4F+B=0\\ \Rightarrow B=-4F\end{array}$

Substituting the value of $B=-4F$ , the equation is reduced to

  $-F{x}^2-4Fxy+C{y}^2+F=0$

Since the point $\left(-2,1\right)$ also lies on the curve, $\left(x=-2\right)$ and $\left(y=1\right)$ should satisfy the above equation. Therefore,

  $\begin{array}{l}-F{\left(-2\right)}^2-4F\left(-2\right)\left(1\right)+C{\left(1\right)}^2+F=0\\ \Rightarrow -4F+8F+C+F=0\\ \Rightarrow C=-5F\end{array}$

Substituting the value of $C=-5F$ , the equation is reduced to

  $-F{x}^2-4Fxy-5F{y}^2+F=0$

Lastly, dividing throughout by $\left(-F\right)$ , the required curve is $x^2+4xy+5y^2-1=0$ .

Conclusion:

Hence, the given conic section is tangent by the curve $x^2+4xy+5y^2-1=0$ .