Question

3x^2-xy+4ky^2=0 is 2x+3y=0​

Answer 1

Answer:

Let another line be y=mx+d

(y−mx−d)(3x+4y=0)

4y2−3mx2+(3−4m)xy−3dx−4yd=0

compare from 

6x2−xy+4cy2=0

d=0

4c4=6−3m=−13−4m

3m=18−24m

27m=18

m=32

c1=6−2⇒c=−3 

Answer 2

Answer:

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