Answer to Question #202772 in Analytic Geometry for tanya

Question #202772

Find the vertices, eccentricity, foci and asymptotes of the hyperbola (x2/8) − (y2 /4) = 1. Also trace it. Under what conditions on λ the line x+λy = 2 will be tangent to this hyperbola? Explain geometrically.


1
Expert's answer
2021-06-12T05:23:24-0400

We compare this equation with x2a2y2b2=1.\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.


Eccentricity is e=1+b2a2=32e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{\dfrac{3}{2}}


The center is C=(0,0)C=(0,0)


The vertices are V=(a,0)=(22,0)V'=(−a,0)=(−2\sqrt{2},0) and V=(a,0)=(22,0)V=(a,0)=(2\sqrt{2},0)


To find the foci, we need the distance from the center to the foci c2=a2+b2=12,  c=±23c^2=a^2+b^2=12, \,\,c=\pm 2\sqrt{3}


The foci are F=(c,0)=(23,0)F'=(-c,0)=(2\sqrt{3},0) and F=(c,0)=(23,0)F=(c,0)=(-2\sqrt{3},0)


The asymptotes are x28y24=0,  y=±12x\dfrac{x^2}{8}-\dfrac{y^2}{4}=0,\,\, y=\pm \dfrac{1}{\sqrt{2}}x





We compare equation of tangent to hyperbola x0xa2y0yb2=1x_0\cdot\dfrac{x}{a^2}-y_0\cdot\dfrac{y}{b^2}=1 with x+λy=2x+\lambda y=2


We have xa2b2y0x0y=a2x0,  x2y0x0y=8x0x-\dfrac{a^2}{b^2} \dfrac{y_0}{x_0}y=\dfrac{a^2}{x_0}, \,\, x-2\dfrac{y_0}{x_0}y=\dfrac{8}{x_0} and x0=4x_0=4, so y0=±2y_0=\pm2 and λ=y02=1\lambda=-\dfrac{y_0}{2}=\mp 1

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