A) 1
B) 2
C) 3
D) 4
Correct Answer: D
Solution :
\[2h=a({{t}_{1}}^{2}+t_{2}^{2})\] ?(i) \[2k=2a({{t}_{1}}+{{t}_{2}})\] ?(ii) and \[{{t}_{1}}{{t}_{2}}=-1\] from (2) \[{{k}^{2}}={{a}^{2}}\left[ \frac{2h}{a}-2 \right]\] \[{{y}^{2}}=2a(x-2a)\] Put \[16x-12y+12=0\] Hence, \[{{y}^{2}}=4(x-4)\]You need to login to perform this action.
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