Hallo, please I assist me understand? (d) Derive the standard equation of the hyperbola with foci along the y-axis and center at (h; k).

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Oct 14, 2017

Answer:

Please see below.

Explanation:

A hyperbola is the set of all points in a plane such that the difference of the distances from two given fixed points called foci is constant.

Let the center of hyperbola be #(h,k)# and focii along #y#-axis be #(h,k-c)# and #(h,k+c)#. Let the difference of the distances be #2a#.

Then from definition we should have

#sqrt((x-h)^2+(y-k+c)^2)-sqrt((x-h)^2+(y-k-c)^2)=2a#

or #sqrt((x-h)^2+(y-k+c)^2)=sqrt((x-h)^2+(y-k-c)^2)+2a#

and squaring both sides we get

#(x-h)^2+(y-k+c)^2=(x-h)^2+(y-k-c)^2+4asqrt((x-h)^2+(y-k-c)^2)+4a^2#

or #y^2+k^2+c^2+2yc-2yk-2kc=y^2+k^2+c^2-2yc-2yk+2kc+4asqrt((x-h)^2+(y-k-c)^2)+4a^2#

or #4yc-4kc=4asqrt((x-h)^2+(y-k-c)^2)+4a^2#

or #asqrt((x-h)^2+(y-k-c)^2)=yc-kc-a^2#

Squaring both sides again, we have

#a^2((x-h)^2+(y-k-c)^2)=(yc-kc-a^2)^2#

or #a^2(x-h)^2+a^2(y-k)^2+a^2c^2color(red)(-2cya^2+2kca^2)=y^2c^2+k^2c^2+a^4color(red)(-2a^2yc+2a^2kc)-2ykc^2#

or #a^2(x-h)^2+a^2(y-k)^2=y^2c^2+k^2c^2+a^4-2ykc^2#

or #a^2(x-h)^2+a^2(y-k)^2=c^2(y^2+k^2-2yk)+a^4#

or #a^2(x-h)^2+a^2(y-k)^2=c^2(y-k)^2+a^4#

or #a^2(x-h)^2-(c^2-a^2)(y-k)^2=a^4#

#(x-h)^2/a^2-(y-k)^2/(a^4/(c^2-a^2))=1#

or #(x-h)^2/a^2-(y-k)^2/(b^2)=1#

where #b^2=a^4/(c^2-a^2)#

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