How do you find the locus of the center of the hyperbola having asymptotes given by $y - x tan α + 1 = 0 and y - x tan (α + \frac{π}{4}) + 2 = 0$ $y-x tan alpha+1=0 and y-x tan(alpha+pi/4)+2=0$ , where $α$ $alpha$ varies?

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How do you find the locus of the center of the hyperbola having asymptotes given by $y - x tan α + 1 = 0 and y - x tan (α + \frac{π}{4}) + 2 = 0$ $y-x tan alpha+1=0 and y-x tan(alpha+pi/4)+2=0$ , where $α$ $alpha$ varies?

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Answer 1 · 2016-12-15T13:50:31

The equations of the locus are:

$x = cos α (cos α - sin α)$ $x=cosalpha(cosalpha-sinalpha)$

$y = - 1 + sin α (cos α - sin α)$ $y=-1+sinalpha(cosalpha-sinalpha)$

Explanation:

By definition the center of the hyperbola is the point where the asymptotes cross, so that they solve the system:

$y - x tan α + 1 = 0$ $y-xtanalpha+1 = 0$
$y - x tan (α + \frac{π}{4}) + 2 = 0$ $y-xtan(alpha+pi/4)+2=0$

$y - x tan α = - 1$ $y-xtanalpha = -1$
$y - x tan (α + \frac{π}{4}) = - 2$ $y-xtan(alpha+pi/4)=-2$

Subtracting member by member:

$x (tan (α + \frac{π}{4}) - tan α) = 1$ $x(tan(alpha+pi/4)-tanalpha)= 1$

So:

$x = \frac{1}{tan (α + \frac{π}{4}) - tan α}$ $x=1/(tan(alpha+pi/4)-tanalpha)$

Now note that:

$\frac{1}{tan (α + \frac{π}{4}) - tan α} = \frac{1}{\frac{sin (α + \frac{π}{4})}{cos (α + \frac{π}{4})} - \frac{sin α}{cos α}} = \frac{cos (α + \frac{π}{4}) cos α}{sin (α + \frac{π}{4}) cos α - sin α cos (α + \frac{π}{4})} = cos α \frac{cos α cos (\frac{π}{4}) - sin α sin (\frac{π}{4})}{sin (α + \frac{π}{4} - α)} = cos α (cos α - sin α)$ $1/(tan(alpha+pi/4)-tanalpha)= 1/(sin(alpha+pi/4)/cos(alpha+pi/4)-sinalpha/cosalpha)= (cos(alpha+pi/4)cosalpha)/(sin(alpha+pi/4)cosalpha-sinalphacos(alpha+pi/4))=cosalpha(cosalphacos(pi/4)-sinalphasin(pi/4))/sin(alpha+pi/4-alpha)=cosalpha(cosalpha-sinalpha)$

so:

$x = cos α (cos α - sin α)$ $x=cosalpha(cosalpha-sinalpha)$

$y = - 1 + tan α cos α (cos α - sin α) = - 1 + sin α (cos α - sin α)$ $y=-1+tanalphacosalpha(cosalpha-sinalpha)=-1+sinalpha(cosalpha-sinalpha)$

Are the equations of the locus.

Answer 2 · 2016-12-15T14:43:48

This locus is the circle, with center at $(\frac{1}{2}, - \frac{3}{2})$ $(1/2, -3/2)$ and
radius $\frac{1}{\sqrt{2}}$ $1/sqrt2$ .

Explanation:

The asymptotes meet at the center of the hyperbola.

Let $t = tan α$ $t = tan alpha$ . Use $tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ $tan(A+B)=(tan A + tan B)/(1-tan A tan B)$ and $tan (\frac{π}{4}) = 1$ $tan(pi/4) = 1$ .

The coordinates (x, y) of the center are given by

$y = x t - 1 = x \frac{t + 1}{1 - t .} - 2$ $y= xt-1=x(t+1)/(1-t.)-2$ . Eliminating t,

$y = x ⎛ ⎜ ⎝ \frac{1 + \frac{y + 1}{x}}{(1 - \frac{y + 1}{x})} ⎞ ⎟ ⎠ - 2$ $y=x((1+(y+1)/x)/((1-(y+1)/x)))-2$

$= x (\frac{x + y + 1}{x - y - 1}) - 2$ $=x((x+y+1)/(x-y-1))-2$ . Cross multiplying,

$x y - y^{2} - y = x^{2} + x y + x - 2 (x - y - 1)$ $xy-y^2-y=x^2+xy+x-2(x-y-1)$ . Simplifying,

$x^{2} + y^{2} - x + 3 y + 2 = 0$ $x^2+y^2-x+3y+2=0$ . In the standard form,

${(x - \frac{1}{2})}^{2} + {(y + \frac{3}{2})}^{2} = {(\frac{1}{\sqrt{2}})}^{2}$ $(x-1/2)^2+(y+3/2)^2=(1/sqrt2)^2$

This locus represents the circle with center at $(\frac{1}{2}, - \frac{3}{2})$ $(1/2, -3/2)$ and

radius $\frac{1}{\sqrt{2}}$ $1/sqrt2$ .

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How do you find the locus of the center of the hyperbola having asymptotes given by $y - x tan α + 1 = 0 and y - x tan (α + \frac{π}{4}) + 2 = 0$ $y-x tan alpha+1=0 and y-x tan(alpha+pi/4)+2=0$ , where $α$ $alpha$ varies?

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Explanation:

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How do you find the locus of the center of the hyperbola having asymptotes given by y-x tan alpha+1=0 and y-x tan(alpha+pi/4)+2=0y−xtanα+1=0andy−xtan(α+π4)+2=0y-x tan alpha+1=0 and y-x tan(alpha+pi/4)+2=0, where alphaαalpha varies?

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Explanation:

Explanation:

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How do you find the locus of the center of the hyperbola having asymptotes given by $y - x tan α + 1 = 0 and y - x tan (α + \frac{π}{4}) + 2 = 0$ $y-x tan alpha+1=0 and y-x tan(alpha+pi/4)+2=0$ , where $α$ $alpha$ varies?