Question #203629

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
2021-06-08T09:36:08-0400
x28y24=1\dfrac{x^2}{8}-\dfrac{y^2}{4}=1

c2=a2+b2=8+4=12c^2=a^2+b^2=8+4=12


Vertices: (a,0),(a,0)(-a, 0), (a, 0)


Vertices: (22,0),(22,0)(-2\sqrt{2}, 0), (2\sqrt{2}, 0)


Eccentricity ca=2322=32=62>1\dfrac{c}{a}=\dfrac{2\sqrt{3}}{2\sqrt{2}}=\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{2}>1


Foci: (c,0),(c,0)(-c, 0), (c, 0)


Foci: (23,0),(23,0)(-2\sqrt{3}, 0), (2\sqrt{3}, 0)


(x)28y24=x28y24\dfrac{(-x)^2}{8}-\dfrac{y^2}{4}=\dfrac{x^2}{8}-\dfrac{y^2}{4}

The curve is symmetrical about the yy -axis.



x28(y)24=x28y24\dfrac{x^2}{8}-\dfrac{(-y)^2}{4}=\dfrac{x^2}{8}-\dfrac{y^2}{4}

The curve is symmetrical about the xx -axis.


The curve is symmetrical in opposite quadrants.


The curve is not symmetrical about the line y=x.y=x.


The origin does not lie on the curve.


yy - axis: x=0=>y24=1.x=0=>-\dfrac{y^2}{4}=1. There is no solution.

There is no yy - intercept.


xx - axis: y=0=>x28=1,x=±22.y=0=>\dfrac{x^2}{8}=1, x=\pm2\sqrt{2}.

yy - intercepts: (22,0),(22,0).(-2\sqrt{2}, 0), (2\sqrt{2}, 0).


Center :(0,0): (0, 0)

First asymptote: y=bax=22xy=-\dfrac{b}{a}x=-\dfrac{\sqrt{2}}{2}x


Second asymptote: y=bax=22xy=\dfrac{b}{a}x=\dfrac{\sqrt{2}}{2}x



First directrix: x=433x=-\dfrac{4\sqrt{3}}{3}


Second directrix: x=433x=\dfrac{4\sqrt{3}}{3}




x28y24=1\dfrac{x^2}{8}-\dfrac{y^2}{4}=1

If λ=0,\lambda =0, then the vertical line x=2x=2 is not tangent to this hyperbola, since the vertices are (22,0),(22,0).(-2\sqrt{2}, 0), (2\sqrt{2}, 0).


The equation of a line is represented by


y=1λx+2λ,λ0y=-\dfrac{1}{\lambda}x+\dfrac{2}{\lambda}, \lambda\not=0

The slope of the tangent at the point (x1,y1)(x_1, y_1) can be obtained by differentiating the equation of the hyperbola.

2x82yy4=0\dfrac{2x}{8}-\dfrac{2yy'}{4}=0

x12y1y(x1,y1)=0x_1-2y_1y'|_{(x_1, y_1)}=0

y(x1,y1)=x12y1y'|_{(x_1, y_1)}=\dfrac{x_1}{2y_1}

The equation of the tangent line is


yy1=x12y1(xx1)y-y_1=\dfrac{x_1}{2y_1}(x-x_1)

y=x12y1xx122y1+y1y=\dfrac{x_1}{2y_1}x-\dfrac{x_1^2}{2y_1}+y_1

y=x12y1xx122y122y1y=\dfrac{x_1}{2y_1}x-\dfrac{x_1^2-2y_1^2}{2y_1}

x128y124=1\dfrac{x_1^2}{8}-\dfrac{y_1^2}{4}=1

x122y12=8x_1^2-2y_1^2=8

The equation of the tangent line is


y=x12y1x82y1y=\dfrac{x_1}{2y_1}x-\dfrac{8}{2y_1}

If 2λ=82y1,\dfrac{2}{\lambda}=-\dfrac{8}{2y_1}, then λ=y14.\lambda=-\dfrac{y_1}{4}.


x12y1=4y1\dfrac{x_1}{2y_1}=-\dfrac{4}{y_1}

x1=8x_1=-8

(8)22y12=8(-8)^2-2y_1^2=8

y1=±27y_1=\pm2\sqrt{7}


Point(8,27),λ=74Point (-8, -2\sqrt{7}), \lambda=\dfrac{\sqrt{7}}{4}



Point(8,27),λ=74Point (-8, 2\sqrt{7}), \lambda=-\dfrac{\sqrt{7}}{4}






Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS