A) \[2x+y=12a\]
B) \[2y+x=12a\]
C) \[2x-y=12a\]
D) \[2y-x=12a\]
Correct Answer: A
Solution :
[a] Equation of tangent of parabola, \[{{y}^{2}}=4ax\] at \[\left( 5a,2a \right)\] be \[y.{{y}_{1}}=2a(x+{{x}_{1}})\] \[2ay=2a\left( x+5a \right)\] \[y=x+5a\] Slope of the tangent, \[m=1\] Slope of the normal \[=-1\] \[\therefore \] Equation of normal of the parabola \[-{{y}^{2}}-4ax\] at \[\left( 5a,2a \right)\] be \[y-2a=\frac{-{{y}_{1}}}{2a}(x-{{x}_{1}})\] \[y-2a=\frac{2a}{2a}(x-{{x}_{1}})\] \[=-1\left( x-5a \right)\] \[x+y=7a\] \[x+y-7a=0\] Hence, option [a] is correct.
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