A) An ellipse of eccentricity\[\frac{2}{\sqrt{3}}\]
B) An ellipse of eccentricity \[\frac{\sqrt{3}}{2}\]
C) Hyperbola of eccentricity 2
D) A hyperbola of eccentricity \[\sqrt{2}\]
Correct Answer: B
Solution :
| \[y=\frac{-2p}{\sqrt{1-{{p}^{2}}}}\operatorname{x}+\frac{1}{\sqrt{1-{{p}^{2}}}};m=\frac{-2p}{\sqrt{1-{{p}^{2}}}}\] |
| \[\Rightarrow \]\[{{m}^{2}}=\frac{4{{p}^{2}}}{1-{{p}^{2}}}\Rightarrow {{m}^{2}}=(4+{{m}^{2}}){{p}^{2}}\] |
| \[\Rightarrow \]\[{{p}^{2}}=\frac{{{m}^{2}}}{4+{{m}^{2}}}\] |
| \[y=mx+\frac{1}{\sqrt{1-\frac{{{m}^{2}}}{4+{{m}^{2}}}}}\] |
| \[\Rightarrow \]\[y=mx+\sqrt{\frac{4+{{m}^{2}}}{4}}\] |
| \[\Rightarrow \]\[y=mx+\sqrt{1+\frac{1}{4}{{m}^{2}}}\] |
| Which touches the ellipse \[\frac{{{x}^{2}}}{1/4}+\frac{{{y}^{2}}}{1}=1\] |
| Whose eccentricity\[e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}\] |
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