A) \[\frac{\sqrt{2}}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{2}{\sqrt{3}}\]
D) \[\frac{2\sqrt{2}}{3}\]
Correct Answer: A
Solution :
Let P (a, b) then equation of normal at P is \[\frac{x}{2a}-\frac{y}{b}=\frac{-1}{2}\] \[\downarrow \left( \frac{-1}{3\sqrt{2}},0 \right)\] \[\Rightarrow \frac{-1}{6\sqrt{2a}}=\frac{-1}{2}\Rightarrow a=\frac{1}{3\sqrt{2}}\] also \[2{{a}^{2}}+{{b}^{2}}=1\] we get \[{{b}^{2}}=1-2\times \frac{1}{18}=\frac{8}{9}\Rightarrow b=\frac{2\sqrt{2}}{3}\] Hence normal is \[\frac{3\sqrt{2x}}{2}-\frac{3y}{2\sqrt{2}}=\frac{-1}{2}\] \[\downarrow (0,\beta )\] \[\beta =\frac{\sqrt{2}}{3}\]
You need to login to perform this action.
You will be redirected in
3 sec
