Question

Equation for parabola of 4y^2+12x-20y+67=0
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Answer 1

Answer:

consider 4y² + 12x −20y + 67 = 0

4 (y²−5y) = −12 x −67

4(y²−2.

2

5

y+

2

2

5

2

)=−12x−67+25=−12x−42