A) \[{{x}^{2}}-{{y}^{2}}-6ax+9{{a}^{2}}=0\]
B) \[{{x}^{2}}-{{y}^{2}}-6ax-6ay+9{{a}^{2}}=0\]
C) \[{{x}^{2}}-{{y}^{2}}-6ay+9{{a}^{2}}=0\]
D) None of these
Correct Answer: A
Solution :
The co-ordinates of the ends of the latus rectum of the parabola \[{{y}^{2}}=4ax\]are \[(a,2a)\]and \[(a,-2a)\] respectively. The equation of the normal at \[(a,2a)\]to \[{{y}^{2}}=4ax\]is \[y-2a=-\frac{2a}{2a}(x-a)\] \[\left[ \text{Using}\,\,y-{{y}_{\text{1}}}=-\frac{{{y}_{1}}}{2a}(x-{{x}_{1}}) \right]\] or \[x+y-3a=0\] .....(i) Similarly, the equation of the normal at \[(a,-2a)\] is \[x-y-3a=0\] ..... (ii) The combined equation of (i) and (ii) is \[{{x}^{2}}-{{y}^{2}}-6ax+9{{a}^{2}}=0\].
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