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Chapter 15: Problem 1453

The latus rectum of the hyperbola \(9 x^{2}-16 y^{2}-18 x-32 y-151=0\) is (a) \((9 / 2)\) (b) \((3 / 2)\) (c) 9 (d) \((9 / 4)\)

Short Answer

Expert verified
The latus rectum of the given hyperbola is \(\frac{9}{2}\).

Step by step solution

01

Rewrite the equation in standard form

To rewrite the given equation \(9x^2 - 16y^2 - 18x - 32y -151 = 0\) in standard form, we'll complete the square for the x and y terms. First, let's group the x and y terms separately: \(9(x^2 - 2x) - 16(y^2 + 2y) = 151\) Now complete the square for x and y terms by adding and subtracting the square of half of the coefficients of x and y inside the parenthesis: \(9[(x-1)^2 - (1)^2] - 16[(y+1)^2 - (1)^2] = 151\) Now, simplifying the equation: \(9(x-1)^2 - 9 - 16(y+1)^2 + 16 = 151\) \[9(x-1)^2 - 16(y+1)^2 = 144\] Now, divide by 144 to get the equation in standard form: \[\frac{(x-1)^2}{16} - \frac{(y+1)^2}{9} = 1\]
02

Find the latus rectum

Now that the equation is in standard form, the latus rectum can be found using the formula for a hyperbola: For the hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the length of the latus rectum is: Latus Rectum = \(2b^2 / a\) In our case, we have the standard form of the hyperbola: \(\frac{(x-1)^2}{16} - \frac{(y+1)^2}{9} = 1\), where \(a^2=16\), and \(b^2=9\). So, \(a=4\) and \(b=3\). Now, we can find the latus rectum: Latus Rectum = \(2b^2 / a = 2(9) / 4 = 18 / 4 = \frac{9}{2}\) Therefore, the latus rectum of the hyperbola is \(\frac{9}{2}\), which corresponds to option (a).

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Key Concepts

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

Standard Form of a Hyperbola
The standard form of a hyperbola is crucial for understanding its properties and for solving related problems. Essentially, it is the form where the equation of a hyperbola is simplified to show its center, vertices, and asymptotes clearly.

A hyperbola with its center at the point \(h, k\) can be expressed in standard form as either \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) for a horizontal hyperbola, or \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) for a vertical hyperbola.

In this representation, \(a\) and \(b\) have geometric significance:
  • \(a\) is the distance from the center to either vertex along the major axis.
  • \(b\) is associated with the distance to the asymptotes and relates to the 'height' of the hyperbola.
To convert the general form of the hyperbola to standard form, we utilize techniques, like 'completing the square.'
Completing the Square
Completing the square is a mathematical method used to transform a quadratic equation into a form that allows for easy identification of its vertex. This technique is also instrumental in rewriting the equation of a hyperbola into standard form.

To complete the square:
  • First, arrange the terms with the same variable together.
  • Divide the coefficient of the first-degree term by 2, then square it.
  • Add and subtract this value within the appropriate grouping of terms.
This process simplifies the hyperbola's equation into separate perfect square terms, and isolates the remaining constant term on one side of the equation. It's an essential skill for analyzing hyperbolas, as it's often the first step in finding properties like vertices and the latus rectum.
Hyperbola Equation Transformation
Hyperbola equation transformation refers to the process of altering the equation’s appearance without changing its geometric properties. The goal is to make the equation more comprehensible or to match a desired form, typically the standard form. In our exercise example, the given hyperbola equation is rewritten by completing the square, which enables the easy extraction of information regarding its shape and size.

After transforming the equation into the standard form, we can conveniently determine various characteristics such as:
  • The center of the hyperbola, where the axes intersect.
  • Orientation, whether it opens horizontally or vertically.
  • The lengths of the axes and the distance to the asymptotes.
  • The latus rectum, which is a segment perpendicular to the transverse axis and passes through a focus.
Understanding the transformation helps students gain deeper insights into the behavior of a hyperbola and how its equation represents its graph.

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Most popular questions from this chapter

The value of \(\mathrm{m}\) for which \(\mathrm{y}=\mathrm{mx}+6\) is a tangent to the hyperbola \(\left(x^{2} / 100\right)-\left(y^{2} / 49\right)=1\) is (a) \(\sqrt{(17 / 20)}\) (b) \(\sqrt{(20 / 3)}\) (c) \(\sqrt{(20 / 17)}\) (d) \(\sqrt{(3 / 20)}\)

Let \(E\) be the ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1\) and \(C\) be the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=9\). Let \(\mathrm{P}\) and \(\mathrm{Q}\) be the point \((1,2)\) and \((2,1)\) respe. Then (a) P lies inside \(\mathrm{C}\) but outside \(\mathrm{E}\) (b) P lies inside both \(\mathrm{C}\) and \(\mathrm{E}\) (c) \(Q\) lies outside both \(\mathrm{C}\) and \(\mathrm{E}\) (d) Q lies inside \(\mathrm{C}\) but outside \(\mathrm{E}\)

A square is inscribed in the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}+4 \mathrm{y}-93=0\) Its sides are parallel to the coordinate axes. Then one vertex of the square is (a) \((1+\sqrt{2},-2)\) (b) \((1-\sqrt{2},-2)\) (c) \((1,-2+\sqrt{2})\) (d) none of these

Two tangents to the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=4\) at the points \(\mathrm{A}\) and \(\mathrm{B}\) meet at \(\mathrm{P}(-4,0)\). The area of the quadrilateral \(\mathrm{PAOB}\), where \(\mathrm{O}\) is the origin is (a) \(4 \sqrt{3}\) (b) 4 (c) \(6 \sqrt{2}\) (d) \(2 \sqrt{3}\)

If tangents to the parabola \(y^{2}=4 \mathrm{ax}\) at the points (at \(_{1}, 2 \mathrm{at}_{1}\) ) and \(\left(\mathrm{at}_{2}^{2}, 2 \mathrm{at}_{2}\right.\) ) intersect on the axis of the parabola, then (a) \(t_{1} t_{2}=-1\) (b) \(\mathrm{t}_{1} \mathrm{t}_{2}=1\) (c) \(\mathrm{t}_{1}=\mathrm{t}_{2}\) (d) \(\mathrm{t}_{1}+\mathrm{t}_{2}=0\)
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