A) 2 units
B) 3 units
C) 4 units
D) 5 units
Correct Answer: C
Solution :
| [c] \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] |
| Hyperbola passes through \[(3\sqrt{5},\,\,1)\] |
| \[\therefore \,\,\,\,\,\frac{{{(3\sqrt{5})}^{2}}}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}=1\] |
| \[\frac{45}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}=1\] ?. (i) |
| Now length of latus rectum \[=\frac{2{{b}^{2}}}{a}\] |
| \[\Rightarrow \frac{4}{3}=\frac{2{{b}^{2}}}{a}\] |
| \[\Rightarrow \frac{2}{3}=\frac{{{b}^{2}}}{a}\Rightarrow a=\frac{3{{b}^{2}}}{2}\] ?. (ii) |
| Putting the value of ?a? for equation (ii) in equation (i), |
| \[\Rightarrow \frac{45\times 4}{9{{b}^{2}}}-\frac{1}{{{b}^{2}}}=1\Rightarrow \frac{20}{{{b}^{4}}}-\frac{1}{{{b}^{2}}}=1\] |
| \[20-{{b}^{2}}={{b}^{4}}\] |
| \[{{b}^{4}}+{{b}^{2}}-20=0\] |
| \[{{b}^{4}}+5{{b}^{2}}-4{{b}^{2}}-20=0\] |
| \[{{b}^{2}}({{b}^{2}}+5)-4({{b}^{2}}+5)=0\] |
| \[({{b}^{2}}-4)({{b}^{2}}+5)=0\] |
| \[{{b}^{2}}=4,{{b}^{2}}=-5\] |
| \[\therefore \,\,\,\,\,\,{{b}^{2}}=4\Rightarrow b=2\] |
| Now length of conjugate axis \[=2b=2(2)=4\] |
You need to login to perform this action.
You will be redirected in
3 sec
