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The parametric equations of a curve are \(x=\sec \theta+1, \quad y=\tan \theta-1\). Its Cartesian equation is: (a) \(y^{2}+3=x^{2}\) (b) \(x^{2}-y^{2}-2 x-2 y=1\) (c) \(x^{2}-y^{2}+2 x+2 y+1=0\) (d) \(x^{2}-y^{2}=1\) (e) \((x-1)^{2}=y^{2}\).

Short Answer

Expert verified
(x - 1)^2 = (y + 1)^2 + 1

Step by step solution

01

Identify the Parametric Equations

The given parametric equations are: \[x = \text{sec}(\theta) + 1\]\[y = \text{tan}(\theta) - 1\]
02

Express \text{sec}(\theta) and \text{tan}(\theta) in Terms of x and y

From the first parametric equation, solve for \( \text{sec}(\theta) \): \[ x - 1 = \text{sec}(\theta) \]From the second parametric equation, solve for \( \text{tan}(\theta) \):\[ y + 1 = \text{tan}(\theta) \]
03

Use the Trigonometric Identity

We know that \[ \text{sec}^2(\theta) - \text{tan}^2(\theta) = 1 \]Substitute the expressions for \( \text{sec}(\theta) \) and \( \text{tan}(\theta) \) into this identity:\[ (x - 1)^2 - (y + 1)^2 = 1 \]
04

Simplify the Equation

Expand and simplify the equation:\[ (x - 1)^2 - (y + 1)^2 = 1 \]The equation is already in a simplified form that matches one of the given options.

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are fundamental tools in mathematics that relate the angles of a triangle to the lengths of its sides. They are especially useful when working with parametric equations. For example, in our exercise, we used the identity \( \text{sec}^2(\theta) - \text{tan}^2(\theta) = 1 \). This identity helps connect different trigonometric functions. Knowing these identities allows us to convert between different forms of equations. Therefore, it's crucial to memorize and understand key trigonometric identities for efficient problem solving.
One practical tip: always express \( \text{sec}(\theta) \) and \( \text{tan}(\theta) \) in terms of x and y, as we did in the solution. This means you can then substitute back into the identity, simplifying the problem substantially.
They can help bridge different mathematical domains, simplifying complex issues.
Cartesian Equation
A Cartesian equation involves x and y coordinates and often describes a curve or line on a plane. It's a crucial representation in mathematics because it allows you to plot equations easily on a graph.
In the exercise, we transformed parametric equations into a Cartesian equation. We started with \(x = \text{sec}(\theta) + 1\) and \(y = \text{tan}(\theta) - 1\), and used trigonometric identities to link x and y directly. Simplifying the equation, we obtained \( (x - 1)^2 - (y + 1)^2 = 1 \). This is now our Cartesian equation.
The importance of having a Cartesian form is that it gives us a more straightforward way to visualize the graph of the curve. Moreover, Cartesian equations are easier to manipulate and integrate into further mathematical operations.
Transformations
Transformations are methods of altering the position or shape of a function's graph. In our problem, we worked on parametric equations that describe a curve. We transformed these into a Cartesian equation. Doing this made the curve easier to understand and visualize.
Common transformations include:
  • Translation: Shifting the graph horizontally or vertically.
  • Rotation: Rotating the graph around a point.
In the context of our problem, the translation helped us align the parametric form to the Cartesian form.
By translating the graph, we set new baselines for x and y coordinates. Then, linking \( x \) and \( y \) through trigonometric identities simplified the transformation process.
Understanding transformations is crucial because it lets us see different perspectives of a graph, enhancing our comprehension and problem-solving skills.

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Most popular questions from this chapter

Find the points of intersection of the circle $$ x^{2}+y^{2}-6 x+2 y-17=0 $$ nd the line \(x-y+2=0\). Show that an equation of the circle which has hese points as the ends of a diameter is $$ x^{2}+y^{2}-4 y-5=0 $$ Show also that this circle and the circle $$ x^{2}+y^{2}-8 x+2 y+13=0 $$ touch externally.

Find the equation of the circle which touches the line \(3 y-4 x-24=0\) at the point \((0,8)\) and also passes through the point \((7,9)\). Prove that this circle also touches the axis of \(x\). Find the equations of the tangents to this circle which are perpendicular to the line \(3 y-4 x-24=0\).

(a) A circle with centre \((3,2)\) touches the line \(4 x-3 y+4=0\). Find the equation of the circle and show that it touches the \(x\)-axis. (b) In each of the following pairs of equations, \(t\) is a parameter. Sketch the locus given by each pair of equations: (i) \(x=3+5 \cos t, \quad y=4+5 \sin t \quad(0 \leqslant t \leqslant \pi)\) (ii) \(x=3 \cos t, \quad y=4 \cos t \quad(0 \leqslant t \leqslant \pi)\) (iii) \(x=3+t \cos \frac{\pi}{3}, \quad y=4+t \sin \frac{\pi}{3} \quad(-\infty

Find the equation of the tangent to the circle \(x^{2}+y^{2}=a^{2}\) at the point \(T(a \cos \theta, a \sin \theta)\). This tangent meets the line \(x+a=0\) at R. If RT is produced to \(\mathrm{P}\) so that \(\mathrm{RT}=\mathrm{TP}\), find the coordinates of \(\mathrm{P}\) in terms of \(\theta\) and find the coordinates of the points in which the locus of \(P\) meets the \(y\)-axis. (U of L)

Obtain the equation of the tangent to the parabola \(y^{2}=4 x\) at the point \(\left(t^{2}, 2 t\right)\). The tangents to the parabola at the points \(\mathrm{P}\left(p^{2}, 2 p\right)\) and \(\mathrm{Q}\left(q^{2}, 2 q\right)\) meet on the line \(y=3\). Find the equation of the locus of the midpoint of \(\mathrm{PQ}\) If \(\mathrm{PQ}\) intersects the \(x\)-axis and the \(y\)-axis at \(\mathrm{R}\) and \(\mathrm{S}\) respectively, find also, the equation of the locus of the midpoint of \(\mathrm{RS}\).

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