question_answer

Locus ot tnsection point ft any arbitrary double ordinate of the parabola \[{{x}^{2}}=4by\], is

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 \[9{{x}^{2}}\,=4by\]