A) 2, 1
B) 1, 2
C) 3, 2
D) 2, 3
Correct Answer: B
Solution :
\[{{y}^{2}}=\pm 4a(x-h)\] Þ \[2y\,{{y}_{1}}=\pm 4a\] Þ \[y{{y}_{1}}=\pm 2a\] Þ \[y_{1}^{2}+y{{y}_{2}}=0\] Hence degree = 1, order = 2.
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