The locus of the point x - t -1- t and y - t -1- t isA- a straight lineB- a circleC- a hyperbolaD- a parabola

The correct option is C a hyperbolaGiven parametric equations are nbsp; x=t+1t ⋯(1) y=t 1t ⋯(2) From (1)^2 (2)^2, we get x^2 y^2=4 On comparing wit ...

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Standard XII

Mathematics

Visualizing Conics from Right Circular Cone

The locus of ...

Question

The locus of the point $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$ is

a straight line

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a circle

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a parabola

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a hyperbola

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Solution

The correct option is D a hyperbola
Given parametric equations are
$x = t + \frac{1}{t} \dots (1) y = t - \frac{1}{t} \dots (2)$
From $(1)^{2} - (2)^{2},$ we get
$x^{2} - y^{2} = 4$
On comparing with $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0,$
we have
$a = 1, h = 0, b = - 1, g = f = 0, c = - 4$
$△ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 4 \neq 0$
and, $h^{2} > a b$
$\Rightarrow$ The locus is a hyperbola.

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The locus of the point x=t+1t and y=t−1t is

The correct option is D a hyperbolaGiven parametric equations are x=t+1t ⋯(1)y=t−1t ⋯(2) From (1)2−(2)2, we get x2−y2=4 On comparing with ax2+2hxy+by2+2gx+2fy+c=0, we have a=1,h=0,b=−1,g=f=0,c=−4 △=abc+2fgh−af2−bg2−ch2=4≠0 and, h2>ab ⇒ The locus is a hyperbola.

The locus of the point $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$ is