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Mathematics Question on Parabola

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

Parabola

Explanation

Solution

Given that:

The set of points of the form(t2+t+1,t2−t+1) (t^2 + t + 1, t^2 - t + 1),

where tt is a real number, represents a parabola.

Let's analyze the given parametric equations:

x=t2+t+1.y=t2−t+1x = t^2 + t + 1.y = t^2 - t + 1

These are parametric equations for the xx and yy coordinates of a point on the plane, where tt is the parameter.

We can eliminate the parameter tt to

Now can express yy in terms of xx, (which will help us identify the geometric shape of the curve.)

Hence we get:

x−y=(t2+t+1)−(t2−t+1)x - y = (t^2 + t + 1) - (t^2 - t + 1)

⇒2tx+y=(t2+t+1)+(t2−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=x−y2t = \dfrac{x - y}{2}

Substitute the expression for tt into x+y:x+y=2(x−y2)2+2x + 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 (x - y)^2 = 2x + 2y - 4 is at the point .(1/2,1/2). (1/2, 1/2). (Hence it can be finally concluded as a parabola for the given question.)