Question

The set of points of the form(t^2+t+1,t^2-t+1),where t is a real number, represents a/an

• A. Circle
• B. Parabola
• C. Ellipse
• D. Hyperbola
• E. pair of straight lines

Answer 1

The Correct Option is B

The set of points of the form\((t^2 + t + 1, t^2 - t + 1)\),

 where \(t\) is a real number, represents a parabola.

Let's analyze the given parametric equations:

\(x = t^2 + t + 1.y = t^2 - t + 1\)

These are parametric equations for the \(x\) and \(y\) coordinates of a point on the plane, where \(t\) is the parameter.

We can eliminate the parameter \(t\) to 

Now can express \(y\) in terms of \(x\), (which will help us identify the geometric shape of the curve.)

Hence we get:

\(x - y = (t^2 + t + 1) - (t^2 - t + 1)\)

\(⇒ 2t x + y = (t^2 + t + 1) + (t^2 - t + 1)\)

                     \(= 2t^2 + 2\)

Now, solving for t in terms of x

 \(t = \dfrac{x - y}{2}\)

Substitute the expression for \(t\) into \(x + y: x + y = 2(\dfrac{x - y}{2})^2 + 2\)

                                                                  ⇒ \(x + y =\dfrac{(x - y)^2}{2} + 2\)

                                                                  ⇒\((x - y)^2 =2x +2y - 4\) 

Now, we have an equation that relates x and y without any parameter t. The equation is a second-degree equation, which represents a parabola.

the vertex of the parabola represented by the equation\((x - y)^2 = 2x + 2y - 4\) is at the point .\((1/2, 1/2).\) 

Answer 2

Step 1: Analyze the given parametric equations.

We are given a set of points in the form:

\[ (x, y) = (t^2 + t + 1, t^2 - t + 1), \]

where \( t \) is a real number. We aim to determine the type of curve represented by these points.

Step 2: Eliminate the parameter \( t \).

From the parametric equations:

\[ x = t^2 + t + 1 \quad \text{and} \quad y = t^2 - t + 1, \]

subtract \( y \) from \( x \):

\[ x - y = (t^2 + t + 1) - (t^2 - t + 1). \]

Simplify:

\[ x - y = 2t. \]

Thus, we can express \( t \) in terms of \( x \) and \( y \):

\[ t = \frac{x - y}{2}. \]

Step 3: Substitute \( t \) back into one of the equations.

Substitute \( t = \frac{x - y}{2} \) into \( x = t^2 + t + 1 \):

\[ x = \left(\frac{x - y}{2}\right)^2 + \frac{x - y}{2} + 1. \]

Simplify each term:

\[ x = \frac{(x - y)^2}{4} + \frac{x - y}{2} + 1. \]

Multiply through by 4 to eliminate fractions:

\[ 4x = (x - y)^2 + 2(x - y) + 4. \]

Expand \( (x - y)^2 \):

\[ 4x = (x^2 - 2xy + y^2) + 2(x - y) + 4. \]

Simplify further:

\[ 4x = x^2 - 2xy + y^2 + 2x - 2y + 4. \]

Rearrange terms:

\[ x^2 - 2xy + y^2 - 2x + 2y + 4 - 4x = 0. \]

Combine like terms:

\[ x^2 - 2xy + y^2 - 6x + 2y + 4 = 0. \]

Step 4: Identify the type of conic section.

The general equation of a conic section is:

\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. \]

Here, the coefficients are:

\[ A = 1, \quad B = -2, \quad C = 1, \quad D = -6, \quad E = 2, \quad F = 4. \]

To classify the conic section, compute the discriminant \( \Delta \):

\[ \Delta = B^2 - 4AC. \]

Substitute \( B = -2 \), \( A = 1 \), and \( C = 1 \):

\[ \Delta = (-2)^2 - 4(1)(1) = 4 - 4 = 0. \]

Since \( \Delta = 0 \), the conic section is a parabola.

Final Answer:

The set of points represents a parabola.