The curve described parametrically by x = t^2 | vedclass

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

Question

(C) Given $x = t^2 + t + 1$ and $y = t^2 - t + 1$. Adding the two equations: $x + y = 2t^2 + 2 = 2(t^2 + 1)$. Subtracting the two equations: $x - y =

Vedclass · Accepted Answer

$A$ parabola

Vedclass · Answer

(C) Given $x = t^2 + t + 1$ and $y = t^2 - t + 1$. Adding the two equations: $x + y = 2t^2 + 2 = 2(t^2 + 1)$. Subtracting the two equations: $x - y = 2t$,which implies $t = \frac{x - y}{2}$. Substituting $t$ into the expression for $x + y$: $x + y = 2\left(\left(\frac{x - y}{2}\right)^2 + 1\right) = 2\left(\frac{(x - y)^2}{4} + 1\right) = \frac{(x - y)^2}{2} + 2$. Multiplying by $2$: $2(x + y) = (x - y)^2 + 4$. Expanding: $2x + 2y = x^2 - 2xy + y^2 + 4$. Rearranging: $x^2 - 2xy + y^2 - 2x - 2y + 4 = 0$. Comparing with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a = 1, h = -1, b = 1$. Since $h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$,the curve represents a parabola.

The curve described parametrically by $x = t^2 + t + 1$ and $y = t^2 - t + 1$ represents:

Similar Questions

The vertex of the parabola $x^2 + 8x + 12y + 4 = 0$ is

If the tangent to the parabola $y^2 = ax$ makes an angle of $45^{\circ}$ with the $x$-axis,what is its point of contact?

If a point $P$ moves such that its distances from the point $A(1, 1)$ and the line $x+y+2=0$ are equal,then the locus of $P$ is

If the angle between the tangents drawn through the point $(-2, -1)$ to the parabola $y^2 = 4x$ is $\theta$,then $\tan 2\theta =$

If $PQ$ is a focal chord of the parabola $y^2=4x$ with focus $S$ and $P=(4,4)$,then $SQ=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module