A) \[{{x}^{2}}+2{{y}^{2}}-ax=0\]
B) \[2{{x}^{2}}+{{y}^{2}}-2ax=0\]
C) \[2{{x}^{2}}+2{{y}^{2}}-ay=0~\]
D) \[2{{x}^{2}}+{{y}^{2}}-2ay=0\]
Correct Answer: B
Solution :
\[T:ty=x+a{{t}^{2}}\] Line perpendicular to (1) through (a, 0) is tx + y = ta ?(2) Equation of OP\[:y-\frac{2}{t}x=0\] ?(3) From eq. (2) and (3) eliminating t we get locusYou need to login to perform this action.
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