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The asymptotes of a hyperbola are y=x+1 and y=3-x. Also, the hyperbola passes through (3,3). Find the distance between the foci of the hyperbola.

Mar 27, 2019

#1
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The asymptotes of a hyperbola are y=x+1 and y=3-x. Also, the hyperbola passes through (3,3). Find the distance between the foci of the hyperbola.

Mar 27, 2019
edited by Omi67  Mar 27, 2019
#2
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The asymptotes of a hyperbola are y=x+1 and y=3-x. Also, the hyperbola passes through (3,3).
Find the distance between the foci of the hyperbola.

Formula hyperbola:

$$\begin{array}{|l|rcll|} \hline 1. \text{ asymptote} & x+1 &=& y \\ & x+1-y &=& 0 \\\\ \hline 2. \text{ asymptote} & 3-x &=& y \\ & 3-x-y &=& 0 \\\\ \hline \text{hyperbola: } & (x+1-y)(3-x-y) + c &=& 0 \\\\ P(x=3,y=3) & (3+1-3)(3-3-3)+c &=& 0 \\ & (1)(-3)+c &=& 0 \\ & c &=& 3 \\ \text{hyperbola: } &\mathbf{ (x+1-y)(3-x-y) + 3} &\mathbf{=}& \mathbf{0} \\\\ & (x+1-y)(3-x-y) + 3 & = & 0 \\ & \Big((x-y)+1\Big)\Big(3-(x+y)\Big) + 3 & = & 0 \\ & 3(x-y)-(x-y)(x+y)+3-(x+y) + 3 & = & 0 \\ & 3(x-y)-(x-y)(x+y)+3-(x+y) + 3 & = & 0 \\ & 3x-3y-(x^2-y^2)+3-x-y + 3 & = & 0 \\ & 2x-4y-x^2+y^2+6 &=& 0 \quad | \quad \cdot (-1) \\ & x^2-2x -y^2+4y -6 &=& 0 \\ & x^2-2x -y^2+4y &=& 6 \\ & (x-1)^2-1 -(y-2)^2+4 &=& 6 \\ & (x-1)^2 -(y-2)^2 &=& 3 \quad | \quad :3 \\ & \dfrac{(x-1)^2}{3} - \dfrac{(y-2)^2}{3} &=& 1 \\ & \boxed{\text{Formula hyperbola:} \dfrac{ (x-h)^2 }{a^2}- \dfrac{ (x-k)^2 }{b^2} = 1 } \\ & a^2 = 3 \\ & a= \sqrt{3} \\\\ & b^2 = 3 \\ & b= \sqrt{3} \\ \hline \end{array}$$

$$\begin{array}{|l|rcll|} \hline \text{Focus}_1: & F_1(h+ae,k) \\ \text{Focus}_2: & F_2(h-ae,k) \\ \hline \text{Foci distance}: & && h+ae-(h-ae) \\ & &=&h+ae-h+ae \\ & &=& 2ae \quad | \quad e=\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=& 2a\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=& 2 \sqrt{a^2+b^2} \quad | \quad a^2=b^2= 3 \\ & &=& 2 \sqrt{3+3} \\ & &\mathbf{=}& 2 \mathbf{ \sqrt{6} } \\ \hline \end{array}$$

The distance between the foci of the hyperbola is $$\mathbf{2\sqrt{6} }$$

Mar 27, 2019
edited by heureka  Mar 27, 2019