The curve described parametrically by x - t 2- t - y -2 t -1 representsA- ParabolaB- An ellipseC- HyperbolaD- Circle

The correct option is A Parabola nbsp;x=t^2+t and y=2t 1 nbsp; By elimination of t nbsp; t=y+12 x=(y+12)^2+y+12 y^2+4y 4x+3 =0 nbsp; On comparing wi ...

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Standard XII

Mathematics

Distinguishing between Conics from General Equation and Eccentricity

The curve des...

Question

The curve described parametrically by $x = t^{2} + t, y = 2 t - 1$ represents

Circle

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Parabola

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An ellipse

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Hyperbola

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Solution

The correct option is B Parabola
$x = t^{2} + t$ and $y = 2 t - 1$

By elimination of $t$

$t = \frac{y + 1}{2}$
$x = {(\frac{y + 1}{2})}^{2} + \frac{y + 1}{2}$

$y^{2} + 4 y - 4 x + 3 = 0$
On comparing with $a x^{2} + b y^{2} + 2 h x y + 2 g x + 2 f y + c = 0$

we have $a = 0, b = 1, g = - 2, f = 2, h = 0, c = 3$
$Δ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = - 4 \neq 0$
and $h^{2} - a b = 0 \Rightarrow h^{2} = a b$
So, the given equation is a parabola

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Distinguishing between Conics from General Equation and Eccentricity

Standard XII Mathematics

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The curve described parametrically by x=t2+t,y=2t−1 represents

The correct option is B Parabola x=t2+t and y=2t−1 By elimination of t t=y+12 x=(y+12)2+y+12 y2+4y−4x+3=0 On comparing with ax2+by2+2hxy+2gx+2fy+c=0 we have a=0,b=1,g=−2,f=2,h=0,c=3 Δ=abc+2fgh−af2−bg2−ch2=−4≠0 and h2−ab=0⇒h2=ab So, the given equation is a parabola

The curve described parametrically by $x = t^{2} + t, y = 2 t - 1$ represents