A) \[xy=\frac{3}{4}\]
B) \[xy=\frac{35}{16}\]
C) \[xy=\frac{64}{105}\]
D) \[xy=\frac{105}{64}\]
Correct Answer: D
Solution :
The given equation of parabola is \[y=\frac{{{a}^{3}}{{x}^{2}}}{3}+\frac{{{a}^{2}}x}{2}-2a\] \[\Rightarrow \]\[y+2a=\frac{{{a}^{3}}}{3}\left( {{x}^{2}}+\frac{3}{2a}x \right)\] \[\Rightarrow \]\[y+2a=\frac{{{a}^{3}}}{3}\left( {{x}^{2}}+\frac{3}{2a}x+\frac{9}{16{{a}^{2}}}-\frac{9}{16{{a}^{2}}} \right)\] \[\Rightarrow \]\[y+2a=\frac{{{a}^{3}}}{3}{{\left( x+\frac{3}{4a} \right)}^{2}}-\frac{9}{16{{a}^{2}}}-\frac{{{a}^{3}}}{3}\] \[\Rightarrow \]\[y+2a+\frac{3a}{16}=\frac{{{a}^{3}}}{3}{{\left( x+\frac{3}{4a} \right)}^{2}}\] \[\Rightarrow \]\[\left( y+\frac{35a}{16} \right)=\frac{{{a}^{3}}}{3}{{\left( x+\frac{3}{4a} \right)}^{2}}\] Thus, the vertices of parabola is\[\left( -\frac{3}{4a},-\frac{35a}{16} \right)\] Let\[h=-\frac{3}{4a}\]and\[k=-\frac{35a}{16}\] Now, \[hk=\frac{105}{64}\] Thus, the locus of vertices of a parabola is \[xy=\frac{105}{64}\]
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