Point of intersection of the 2 curves
#x^2=16y#
#y=x^2/16#
Sub #y=x^2/16# into #xy=4#
#xtimesx^2/16=4#
#x^3/16=4#
#x^3=64#
#x=4#
Sub #x=4# into #y=x^2/16#
#y=4^2/16#
#y=1#
Therefore, #(4,1)# is the point of intersection.
The angle between 2 points can be found using the formula
#tan theta=abs((m_1-m_2)/(1+m_1m_2))#
where #m_1# and #m_2# are the gradients of the lines
To find the gradients of the line when #x=4#, we need to find the first derivative of both lines and sub #x=4# into each
#xy=4#
#y=4/x#
#(dy)/(dx)=-4/x^2#
#(dy)/(dx)=-4/(4^2)#
#m_1=(dy)/(dx)=-1/4#
#x^2=16y#
#y=x^2/16#
#(dy)/(dx)=x/8#
#(dy)/(dx)=4/8#
#m_2=(dy)/(dx)=1/2#
#tantheta=abs((-1/4-1/2)/(1+(-1/4times1/2)))#
#tantheta=abs((-3/4)/(7/8))#
#tantheta=abs(-6/7)#
#tantheta=6/7#
#theta=tan^(-1) (6/7)#
#theta=40'36"#
