A) \[\left( \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}} \right)=\frac{{{p}^{2}}{{x}^{2}}}{{{a}^{4}}}+\frac{{{q}^{2}}{{y}^{2}}}{{{b}^{4}}}\]
B) \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=\frac{{{x}^{2}}}{{{p}^{2}}}+\frac{{{y}^{2}}}{{{q}^{2}}}\]
C) \[\left( \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}} \right)=\frac{{{p}^{2}}}{{{a}^{2}}}+\frac{{{q}^{2}}}{{{b}^{2}}}\]
D) \[\left( \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}} \right)=\frac{{{p}^{2}}}{{{q}^{2}}}+\frac{{{a}^{2}}}{{{b}^{2}}}\]
Correct Answer: A
Solution :
Let the mid-point of any chord be (h, k). Then, \[{{k}_{2}}\times {{10}^{3}}={{k}_{1}}\] \[+2{{H}_{2}}O+N{{H}_{4}}Cl\] \[-OC{{H}_{3}}\] \[-I\] \[OC{{H}_{3}}\] \[A{{l}_{2}}{{S}_{3}}+6{{H}_{2}}O\xrightarrow{\,}\,2Al{{(OH)}_{3}}+3{{H}_{2}}S\] \[M{{g}^{2+}}\] \[A{{l}^{3+}}\] Condition for tangency between the line \[=2\times 4=8\] and ellipse is \[\therefore \] \[=\frac{1}{8}\times 100\] \[AlC{{l}_{3}}\] Hence, locus of a point is \[HCl\] \[AlC{{l}_{3}}+3{{H}_{2}}O\xrightarrow{\,}\,Al{{(OH)}_{3}}+3HCl\] \[(a=b=c\] \[\alpha =\beta =\gamma ={{90}^{o}}\] \[(a=b=c\]
You need to login to perform this action.
You will be redirected in
3 sec
