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Chapter 10: Problem 36

Prove: If \(B \neq 0,\) then the graph of \(x^{2}+B x y+F=0\) is a hyperbola if \(F \neq 0\) and two intersecting lines if \(F=0\)

Short Answer

Expert verified
The graph is a hyperbola if \(F \neq 0\) and two intersecting lines if \(F = 0\).

Step by step solution

01

Check the Equation Form

Write down the given equation: \(x^2 + Bxy + F = 0\). This equation appears to be a conic section represented by a general quadratic equation of the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) where \(A = 1\), \(C = 0\), \(D = 0\), \(E = 0\), and we have the \(B\) coefficient affecting the \(xy\) term.
02

Evaluate the Discriminant

For conic sections, the discriminant \(\Delta = B^2 - 4AC\) determines the type of the curve. Plug in the values: \(\Delta = B^2 - 4 \times 1 \times 0 = B^2\). Since \(B eq 0\), \(\Delta = B^2 > 0\). When the discriminant is greater than zero, the conic section is a hyperbola.
03

Analyze the Case \(F = 0\)

For \(F = 0\), the equation reduces to \(x^2 + Bxy = 0\). Factorize this equation: \(x(x + By) = 0\). The solutions are \(x = 0\) or \(x + By = 0\), which represent two intersecting lines through the origin.
04

Analyze the Case \(F \neq 0\)

With \(F eq 0\), the equation \(x^2 + Bxy + F = 0\) cannot be factorized into linear factors, reaffirming that it represents a hyperbola as previously found from the discriminant being positive.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hyperbola
A hyperbola is a type of conic section that looks like two mirrored curves. It can be thought of as a set of points where the difference in distances between each point and two fixed points, called foci, is constant.

In terms of algebra, a hyperbola is recognized by its general equation form:
  • When you see a quadratic equation like \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), the type of conic section can be identified using the discriminant.
  • The discriminant \(\Delta = B^2 - 4AC\) helps in identifying the specific conic: when \(\Delta > 0\), it indicates a hyperbola.
  • In our exercise equation, \(x^2 + Bxy + F = 0\), the discriminant simplifies to \(B^2\), as \(C = 0\), leading to \(\Delta = B^2\).
Since \(B eq 0\), \(\Delta > 0\), confirming the presence of a hyperbola when \(F eq 0\). This nature of the equation aligns with the property of a hyperbola being formed with intersecting axes, originating across different sides of a center point.
Discriminant
The discriminant is a crucial determinant in classifying conic sections. It provides insight into the geometric nature of the equation represented.
When dealing with a quadratic equation of the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), the discriminant is calculated as \(\Delta = B^2 - 4AC\). Here's how it works:
  • If \(\Delta > 0\), the equation represents a hyperbola.
  • If \(\Delta = 0\), it indicates a parabola.
  • If \(\Delta < 0\), it corresponds to an ellipse or a circle.
In the specific exercise presented, the equation \(x^2 + Bxy + F = 0\) makes use of the discriminant in determining that \(\Delta = B^2\), since \(A = 1\) and \(C = 0\). Hence, a positive \(\Delta\) assures us that what we are dealing with is indeed a hyperbola, and \(F eq 0\) further reinforces this geometric nature.
Intersecting Lines
Intersecting lines are formed when the equation can break down into two linear factors. This happens when the constant term allows perfect factorization.

Consider the equation \(x^2 + Bxy + F = 0\). Let's explore the condition for intersecting lines:
  • When \(F = 0\), the equation simplifies to \(x^2 + Bxy = 0\). This can be factorized as \(x(x + By) = 0\).
  • Solving each factor gives the lines \(x = 0\) and \(x + By = 0\), which are two intersecting lines passing through the origin.
These lines meet at the origin and form a set of intersecting lines, illustrating how an equation can shift its geometric interpretation by the value of coefficents.
Intersecting at the point, usually the origin, gives a visual representation as simple straight lines crossing each other on a graph.

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Most popular questions from this chapter

Show that in a polar coordinate system the distance \(d\) between the points \(\left(r_{1}, \theta_{1}\right)\) and \(\left(r_{2}, \theta_{2}\right)\) is $$ d=\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)} $$

Show that the curve with parametric equations $$ x=t^{2}-3 t+5, \quad y=t^{3}+t^{2}-10 t+9 $$ intersects itself at the point \((3,1),\) and find equations for the two tangent lines to the curve at the point of intersection.

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5. ] Use the formulas above in these exercises. Find the area of the surface generated by revolving \(x=6 t\) \(y=4 t^{2}(0 \leq t \leq 1)\) about the \(y\) -axis.

True鈥揊alse Determine whether the statement is true or false. Explain your answer. $$ \begin{array}{l}{\text { The area of a sector with central angle } \theta \text { taken from a circle }} \\ {\text { of radius } r \text { is } \theta r^{2} .}\end{array} $$

Find a polar equation for the conic that has its focus at the pole and satisfies the stated conditions. Points are in polar coordinates and directrices in rectangular coordinates for simplicity. (In some cases there may be more than one conic that satisfies the conditions.) $$ \begin{array}{l}{\text { (a) Ellipse; ends of major axis }(2, \pi / 2) \text { and }(6,3 \pi / 2) .} \\ {\text { (b) Parabola; vertex }(2, \pi) .} \\\ {\text { (c) Hyperbola; } e=\sqrt{2} ; \text { vertex }(2,0)}\end{array} $$
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