A) \[\frac{11}{3}\]
B) \[-\frac{11}{3}\]
C) \[\frac{13}{2}\]
D) \[-\frac{13}{2}\]
Correct Answer: D
Solution :
Let the coordinate at point of intersection of normals at P and Q be (h, k) Since, equation of normals to the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]At point \[({{x}_{1}},{{y}_{1}})\]is\[\frac{{{a}^{2}}x}{{{x}_{1}}}+\frac{{{b}^{2}}y}{{{y}_{1}}}={{a}^{2}}+{{b}^{2}}\]therefore equation of normal to the hyperbola \[\frac{{{x}^{2}}}{{{3}^{2}}}-\frac{{{y}^{2}}}{{{2}^{2}}}=1\] at point\[P(3sec\theta ,2tan\theta )\]is \[\frac{{{3}^{2}}x}{3\sec \theta }+\frac{{{2}^{2}}y}{2\tan \theta }={{3}^{2}}+{{2}^{2}}\] \[\Rightarrow \]\[\] ?.(1) Similarly, Equation of normal to the hyperbola \[\frac{{{x}^{2}}}{{{3}^{2}}}-\frac{{{y}^{2}}}{{{2}^{2}}}\]at point \[Q(3sec\theta ,2tan\theta )\]is \[\frac{{{3}^{2}}x}{3\sec \phi }+\frac{{{2}^{2}}y}{2tan\phi }={{3}^{2}}+{{2}^{2}}\] \[\Rightarrow \]\[\] ?.(2) Given \[\theta +\phi =\frac{\pi }{2}\Rightarrow \phi =\frac{\pi }{2}-\theta \]and these passes through (h, k) \[\therefore \]From eq. (2) \[3x\cos \left( \frac{\pi }{2}-\theta \right)+2y\cot \left( \frac{\pi }{2}-\theta \right)={{3}^{2}}+{{2}^{2}}\] \[\Rightarrow \]\[\] ?(3) and \[\] ...(4) Comparing equation (3) & (4), we get \[3h\cos \theta +2k\cot \theta =3h\sin \theta 2k\tan \theta \] \[3h\cos \theta -3k\sin \theta =2k\tan \theta -2k\cot \theta \] \[3h(\cos \theta -\sin \theta )=2k(\tan \theta -\cot \theta )\] \[3h(\cos \theta -\sin \theta )=2k\frac{(sin\theta -\cos \theta )(sin\theta +\cos \theta )}{sin\theta \cos \theta }\]or,\[3h=\frac{-2k(sin\theta +\cos \theta )}{sin\theta \cos \theta }\] ...(5) Now, putting the value of equation (5) in eq. (3)\[\frac{-2k(sin\theta +\cos \theta )sin\theta }{\cancel{sin}\theta \cos \theta }+2k\tan \theta ={{3}^{2}}+{{2}^{2}}\] \[\Rightarrow \]\[2k\cancel{\tan }\theta -2k+2k\cancel{\tan }\theta =13\] \[-2k=13\Rightarrow k=\frac{-13}{2}\] Hence, ordinate of point of intersection of normals at P and Q is\[\frac{-13}{2}\]
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