Observe that coefficients of #x^2# and #y^2# are opposite in sign.
Hence, this can be written as #7x^2-5y^2=35# and dividing all terms by #35#, we get
#x^2/5-y^2/7=1#, which is the equation of hyperbola with centre at #(0,0)#. Observe that when #y-0#, #x^2=5# andhence intercepts on #x#-axis are #-sqrt5# and #sqrt5#. Observe that if #x# lies between these two value, #y# is not defined as #y^2 < 0#.
Also observe that if #(x,y)# lies oncurve than #(x,-y)#, #(-x,-y)# and #(-x,y)# too lie on the graph and hence graph is symmetric with respect to #x# and #y# axes.
Other points can be identified by putting values of #x# less than #-sqrt5# and greater than #sqrt5#, say #x={-5,-4,-3,3,4,5}# for which we get #y={+-2sqrt7,+-sqrt(77/5),+-2sqrt(7/5),+-+-2sqrt(7/5),sqrt(77/5)+-2sqrt7}# or #{+-5.29,+-3.92,+-2.37,+-2.37,+-3.92,+-5.29}#.
The graph appears as follows:
graph{x^2/5-y^2/7=1 [-10, 10, -5, 5]}
