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.
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Expert's answer
2021-06-08T09:36:08-0400
8x2−4y2=1
c2=a2+b2=8+4=12
Vertices: (−a,0),(a,0)
Vertices: (−22,0),(22,0)
Eccentricity ac=2223=23=26>1
Foci: (−c,0),(c,0)
Foci: (−23,0),(23,0)
8(−x)2−4y2=8x2−4y2
The curve is symmetrical about the y -axis.
8x2−4(−y)2=8x2−4y2
The curve is symmetrical about the x -axis.
The curve is symmetrical in opposite quadrants.
The curve is not symmetrical about the line y=x.
The origin does not lie on the curve.
y - axis: x=0=>−4y2=1. There is no solution.
There is no y - intercept.
x - axis: y=0=>8x2=1,x=±22.
y - intercepts: (−22,0),(22,0).
Center :(0,0)
First asymptote: y=−abx=−22x
Second asymptote: y=abx=22x
First directrix: x=−343
Second directrix: x=343
8x2−4y2=1
If λ=0, then the vertical line x=2 is not tangent to this hyperbola, since the vertices are (−22,0),(22,0).
The equation of a line is represented by
y=−λ1x+λ2,λ=0
The slope of the tangent at the point (x1,y1) can be obtained by differentiating the equation of the hyperbola.
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