A) The length of the transverse axes is 4.
B) Length of latus rectum is 9.
C) Equation of directrix is \[x=\frac{21}{5}\] and \[x=-\frac{11}{5}\]
D) None of these
Correct Answer: C
Solution :
| We have, \[9{{x}^{2}}-16{{y}^{2}}-18x+32y-151=0\] |
| \[\Rightarrow \,\,\,9({{x}^{2}}-2x)-16({{y}^{2}}-2y)=151\] |
| \[\Rightarrow \,\,9({{x}^{2}}-2x+1)-16({{y}^{2}}-2y+1)=144\] |
| \[\Rightarrow \,\,9{{(x-1)}^{2}}-16{{(y-1)}^{2}}=144\] |
| \[\frac{{{(x-1)}^{2}}}{16}-\frac{{{(y-1)}^{2}}}{9}=1\] |
| Shifting the origin at \[(1,1)\] without rotating the axes \[\frac{{{X}^{2}}}{16}-\frac{{{Y}^{2}}}{9}=1\] where \[x=X+1\] and \[y=Y+1\] |
| This is of the form \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] where \[{{a}^{2}}=16\] and \[{{b}^{2}}=9\] so the length of the transverse axes \[=2a=8\] |
| The length of the latus rectum \[=\frac{2{{b}^{2}}}{a}=\frac{9}{2}\] |
| The equation of the directrix is \[X=\pm \frac{a}{e}\] |
| \[x-1=\pm \frac{16}{5}\left( \because \,e=\frac{5}{4} \right)\] |
| \[\Rightarrow \,\,x=\pm \frac{16}{5}+1\Rightarrow x=\frac{21}{5}\] and \[x=-\frac{11}{5}\] |
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