Question

y = px+ root(a^2p^2 +p^2) find singular solution
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Answer 1

Answer:

y²+p²-2pyx=a²(1+p²)

a²p²-p²x²+2pyx+a²-y²=0

p²(a²-x²)+p(2yx)+(a²-y²)=0

D=(2yx)²-(a²-x²)(a²-y²)=0

4(xy)²-4(a²-x²)(a²-y²)=0

x²y²-(a⁴)

ax²+6x+6=0

D=b²-4ac=0