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If a!=0 and the line 2bx+3cy+4d=0 passes...

If `a!=0` and the line `2bx+3cy+4d=0` passes through the points of intersection of the parabola `y^2 = 4ax` and `x^2 = 4ay`, then

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Intersection points of `y^2 = 4ax` and `x^2 = 4ay` will be `(0,0)` and `(4a,4a)`.
Putting these values of `(x,y)` in given line, `2bx+3cy+4d = 0`
At point `(0,0),`
`0+0+4d = 0 => d = 0->(1)`
At point `(4a,4a),`
` 8ab+12ac+4ad = 0`
`=>4a(2b+3c+0) = 0`
`=>4a(2b+3c) = 0`
As ,` a !=0,` So, `2b+3c = 0->(2)`
Squaring and adding (1) and (2),
...
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