A) \[8\]
B) \[10\]
C) \[12\]
D) \[15\]
Correct Answer: B
Solution :
[b] Given line is \[4ax+3by=24\] ...(1) Let this line be normal to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] at \[(a\,\,\cos \theta ,\,\,b\sin \theta )\]. So, equation of normal to ellipse at \[(a\,\,\cos \theta ,\,\,b\sin \theta )\]is: \[ax\,\,\sec \theta -by\,\cos ec\,\theta =({{a}^{2}}-{{b}^{2}})\] ... (2) Equations (1) and (2) are identical, so \[\frac{\sec \theta }{4}=\frac{-\cos ec\,\theta }{3}=\frac{{{a}^{2}}-{{b}^{2}}}{24}\] \[\therefore \,\,\,\,\,\,\cos \theta =\frac{6}{{{a}^{2}}-{{b}^{2}}}\] and \[sin\theta =\frac{-8}{{{a}^{2}}-{{b}^{2}}}\] Squaring and adding, we get \[\frac{36}{{{({{a}^{2}}-{{b}^{2}})}^{2}}}+\frac{64}{{{({{a}^{2}}-{{b}^{2}})}^{2}}}=1\] \[\Rightarrow \,\,\,\,\,100={{({{a}^{2}}-{{b}^{2}})}^{2}}\]\[\Rightarrow \,\,\,\,\,({{a}^{2}}-{{b}^{2}})=10\,\,\,\,\,\,\,\,(\because \,\,a>b)\]
You need to login to perform this action.
You will be redirected in
3 sec
