How do you solve #x^2+y^2=25# and #y+5=1/2x^2 # using substitution?

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Answer 1 · 2016-04-04T10:09:23

Solution set for #(x,y)# is #(0,-5)#, #(4,3)# and #(-4,3)#

As #1/2x^2=y+5#, we have #x^2=2*(y+5)=2y+10#.

Putting this in #x^2+y^2=25#, we get #2y+10+y^2=25# or

#y^2+2y-15=0# i.e. #y^2+5y-3y-15=0# or

#y(y+5)-3(+5)=0# or #(y-3)(y+5)=0# i.e. #y=3# or #y=-5#

Hence #x^2=2*3+10=16# i.e. #x=+-4# that is

#x^2=2*(-5)+10=0# i.e. #x=0#

Hence solution set for #(x,y)# is #(0,-5)#, #(4,3)# and #(-4,3)#

Answer 2 · 2016-04-04T10:31:45

#color(green)("Points of intersection are:")#

#color(blue)((x,y)->( -4,3)" ; "(+4,3)" ; "(0,-5))#

Just an observation: #x^2+y^2=25# is the equation of a circle with the centre at the origin.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:
#x^2+y^2=25# ...............................(1)
#y+5=1/2 x^2# ................................(2)

#color(blue)("Solving for "y)#

Rewrite (1) as: #x^2=25-y^2" " ...........(1_a)#
Rewrite (2) as: #x^2=2y+10" " ...........(2_a)#

Equate #(1_a)" to "(2_a)" through "x^2# This is the equivalent of substituting for #x^2#

#25-y^2=2y+10#

This is the same as

#y^2+2y-15=0#

#(y+5)(y-3)=0#

#" "color(blue)(=> y= -5" or " +3)# .........................(3)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solving for "x)#

Substituting into equation #(2_a)#. The y is not squared!

#color(brown)("condition 1 "y=-5)#

#x^2=2y+10" " -> " "x^2=2(-5)+10#

#" "color(green)(x^2=0" " =>" " x=0)#
'...........................................
#color(brown)("condition 2 "y=+3)#

#x^2=2(3)+10 = 16#

#" "color(green)(x=+-4)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#

The point of intersection are:

#color(blue)((x,y)->( -4,3)" ; "(+4,3)" ; "(0,-5))#

Tony B