What does the curve defined parametrically by | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given,$x = t^2 + t + 1 \quad (i)$ and $y = t^2 - t + 1 \quad (ii)$.Adding $(i)$ and $(ii)$,we get $x + y = 2t^2 + 2 = 2(t^2 + 1) \quad (iii)$.Subt

Vedclass · Accepted Answer

$A$ parabola

Vedclass · Answer

(C) Given,$x = t^2 + t + 1 \quad (i)$ and $y = t^2 - t + 1 \quad (ii)$.Adding $(i)$ and $(ii)$,we get $x + y = 2t^2 + 2 = 2(t^2 + 1) \quad (iii)$.Subtracting $(ii)$ from $(i)$,we get $x - y = 2t$,which implies $t = \frac{x - y}{2} \quad (iv)$.Substituting $(iv)$ into $(iii)$:$x + y = 2 \left[ \left( \frac{x - y}{2} \right)^2 + 1 \right]$$x + y = 2 \left[ \frac{x^2 + y^2 - 2xy}{4} + 1 \right]$$x + y = \frac{x^2 + y^2 - 2xy}{2} + 2$$2x + 2y = x^2 + y^2 - 2xy + 4$$x^2 - 2xy + y^2 - 2x - 2y + 4 = 0$This is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,where $a=1, h=-1, b=1, g=-1, f=-1, c=4$.The discriminant $h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$.Since $h^2 - ab = 0$,the equation represents a parabola.

What does the curve defined parametrically by $x = t^2 + t + 1$ and $y = t^2 - t + 1$ represent?

Similar Questions

Find the equation of the parabola that satisfies the following conditions: Vertex $(0, 0)$,focus $(-2, 0)$.

The equation $y^{2}+4x+4y+k=0$ represents a parabola whose latus rectum is

Let $O$ be the vertex of the parabola $y^2 = 4x$ and its chords $OP$ and $OQ$ are perpendicular to each other. If the locus of the mid-point of the line segment $PQ$ is a conic $C$, then the length of its latus rectum is:

$A$ circle of radius $4$,drawn on a chord of the parabola $y^2 = 8x$ as diameter,touches the axis of the parabola. Then,the slope of the chord is

The length of the chord of the parabola $x^2 = 4y$ having the equation $x - \sqrt{2}y + 4\sqrt{2} = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module