How do you find all the critical points to graph #9x^2 – y^2 – 36x + 4y + 23 = 0 # including vertices, foci and asymptotes?

2 Answers
Jul 18, 2016

Answer:

A hyperbole #9(x-2)^2-(y-2)^2-9=0# with asymptotes
#(x-2)+1/3(y-2)=0# and
#(x-2)-1/3(y-2)=0#

Explanation:

This conic can be represented formally with the structure

#h(x,y)=(x-x_0)^2/a-(y-y_0)^2/b+c=0#

and is characterized as a hyperbole.

Forcing

#9 x^2 - y^2 - 36 x + 4 y + 23-h(x,y)=0, forall {x,y} in RR^2#

we have the conditions

#{ (c + x_0^2/a - y_0^2/b -23= 0), ( (2 y_0)/b-4 = 0), ( 1 - 1/b = 0), (36 - (2 x_0)/a = 0), (1/a -9= 0) :}#

Solving for #a,b,c,x_0,y_0# we obtain

#{a = 1/9, b = 1, c = -9, x_0 = 2, y_0 = 2}#

characterizing #h(x,y)# as

#9(x-2)^2-(y-2)^2-9=0#

Also #h(x,y)# can be written as

#(sqrt(b)(x-x_0)+sqrt(a)(y-y_0))(sqrt(b)(x-x_0)-sqrt(a)(y-y_0))+a b c=0#

The two lines

#sqrt(b)(x-x_0)+sqrt(a)(y-y_0)=0# and
#sqrt(b)(x-x_0)-sqrt(a)(y-y_0)=0# are the asymptotes, or

#(x-2)+1/3(y-2)=0# and
#(x-2)-1/3(y-2)=0#

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Jul 18, 2016

Answer:

#(1)# Vertices : #(3,2), (1,2), (2,5), (2,-1)#

#(2)# Focii : #(2+sqrt10,2), (2-sqrt10,2)#

#(3)# Eqns. of Asymptotes : #3x-y-4=0, 3x+y-8=0#

Explanation:

We complete the squares in the given eqn. to get the modified eqn.

#9x^2-36x+36-y^2+4y-4=-23+36-4#

#:.9(x^2-4x+4)-(y^2-4y+4)=9#

#:.9(x-2)^2-(y-2)^2=9#

#:. (x-2)^2/1-(y-2)^2/9=1#

Letting, #x-2=X, y-2=Y#, we have,

#X^2/1-Y^2/9=1#, which represents a Hyperbola .

Comparing with, #X^2/a^2-Y^2/b^2=1#, we have,

#a^2=1,b^2=9#

Hence the Vertices, in #(X,Y)# system, are #(+-a,0)=(+-1,0) and

(0,+-b)=(0,+-3)#

In #(x,y)#, they are, #x-2=+-1,y-2=0#, i.e., #x=2+-1,y=2#, or, #(3,2), &,

(1,2)#.

#(2,2+-3), i.e., (2,5) and (2,-1)# are also vertices.

The Eccentricity #e# is given by,

#b^2=a^2(e^2-1) rArr 9=e^2-1rArre^2=10rArre=sqrt10#.

Focii , in #(X,Y)# system, are #(+-ae,0)=(+-sqrt10,0)#

#:.#, In #(x,y)# system, Focii are given by, #x-2=+-sqrt10, y-2=0# i.e., 3x=2+-sqrt10, y=2#.

#:.# Focii, in original system, are #(2+-sqrt10,2)#

Asymptotes in #(X,Y)# are, #X^2/1-Y^2/9=0#, i.e., #Y=+-3X#.

So, in #(x,y)#, they are, #y-2=+-3(x-2)# or, #3x-y-4=0, 3x+y-8=0#

Altogether,

#(1)# Vertices : #(3,2), (1,2), (2,5), (2,-1)#

#(2)# Focii : #(2+sqrt10,2), (2-sqrt10,2)#

#(3)# Eqns. of Asymptotes : #3x-y-4=0, 3x+y-8=0#