A) \[ty=x{{t}^{2}}+a\]
B) \[ty=x+a{{t}^{2}}\]
C) \[y=tx+a{{t}^{2}}\]
D) \[y=tx+(a/{{t}^{2}})\]
Correct Answer: A
Solution :
Equation of the tangent to the parabola, \[{{y}^{2}}=4ax\] is \[y{{y}_{1}}=2a(x+{{x}_{1}})\] Þ \[y.\frac{2a}{t}=2a\left( x+\frac{a}{{{t}^{2}}} \right)\] Þ \[\frac{y}{t}=\left( x+\frac{a}{{{t}^{2}}} \right)\ Þ\[\frac{y}{t}=\frac{{{t}^{2}}x+a}{{{t}^{2}}}\] Þ \[ty={{t}^{2}}x+a\]
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