A) \[\frac{x}{3}-\frac{y}{2}=1,\,\,\frac{x}{2}+\frac{y}{3}=\frac{5}{6}\]
B) \[\frac{x}{3}+\frac{y}{2}=1,\,\,\frac{x}{2}-\frac{y}{3}=\frac{5}{6}\]
C) \[\frac{x}{2}+\frac{y}{3}=1,\,\,\frac{x}{3}-\frac{y}{2}=\frac{5}{6}\]
D) None of the above
Correct Answer: A
Solution :
Given equation of ellipse is\[4{{x}^{2}}+9{{y}^{2}}=36\] or\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1\] Tangent at point\[(3,-2)\] is\[\frac{(3)x}{9}+\frac{(-2)y}{4}=\]or\[\frac{x}{3}-\frac{y}{2}=1\] \[\therefore \]Normal is\[\frac{x}{2}+\frac{y}{3}=k\]and it passes through point\[(3,-2)\]. Then,\[\frac{3}{2}-\frac{2}{3}=k\Rightarrow k=\frac{5}{6}\] Hence, normal is \[\frac{x}{2}+\frac{y}{3}=\frac{5}{6}\] and tangent is\[\frac{x}{3}-\frac{y}{2}=1\].
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