A) c
B) 2c
C) ? c
D) 3c
Correct Answer: A
Solution :
Tangent at \[{{y}^{2}}=4ax\]is \[y=mx+\frac{a}{m}\] Therefore, tangent at \[{{y}^{2}}=4a(x+a)\] is, \[y=m(x+a)+\frac{a}{m}\] or \[y=mx+ma+\frac{a}{m}\]Þ\[ma+\frac{a}{m}=c\].You need to login to perform this action.
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