A) \[9{{x}^{2}}=by\]
B) \[3{{x}^{2}}=2by\]
C) \[9{{x}^{2}}=4by\]
D) \[9{{x}^{2}}=2by\]
Correct Answer: C
Solution :
Let \[A\equiv \,(2bt,\,\,b{{t}^{2}})\,B\equiv \,(-2bt,\,\,b{{t}^{2}})\] be the extremities on the double, ordinate AB. If C (h, k) be it?s trisection point, then \[6h=4bt\Rightarrow \,k=b{{t}^{2}}\] \[\Rightarrow \,\,\,t=\frac{3h}{2b},\,{{t}^{2}}=\frac{k}{b}\Rightarrow \,\frac{k}{b}=\frac{9{{h}^{2}}}{4{{b}^{2}}}\] Thus locus of C is \[9{{x}^{2}}\,=4by\]You need to login to perform this action.
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