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

Updated On: May 29, 2024
  • Circle

  • Parabola

  • Ellipse

  • Hyperbola

  • pair of straight lines

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The Correct Option is B

Solution and Explanation

Given that:

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 .\( (1/2, 1/2).\) (Hence it can be finally concluded as a parabola for the given question.)

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Concepts Used:

Parabola

Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).

Parabola


 

 

 

 

 

 

 

 

 

Standard Equation of a Parabola

For horizontal parabola

  • Let us consider
  • Origin (0,0) as the parabola's vertex A,
  1. Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
  2. P(x,y) as the moving point.
  • Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
  • The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
  • The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
  • By definition PM = PS

=> MP2 = PS2 

  • So, (a + x)2 = (x - a)2 + y2.
  • Hence, we can get the equation of horizontal parabola as y2 = 4ax.

For vertical parabola

  • Let us consider
  • Origin (0,0) as the parabola's vertex A
  1. Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
  2. P(x,y) as any moving point
  • Let us now draw a perpendicular SZ from S to the directrix.
  • Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
  • The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
  • By definition PM = PS

=> MP2 = PS2

So, (b + y)2 = (y - b)2 + x2

  • As a result, the vertical parabola equation is x2= 4by.