A) 4ab
B) \[\frac{ab\sqrt{{{a}^{2}}+{{b}^{2}}}}{4}\]
C) \[\frac{ab\sqrt{{{a}^{2}}-{{b}^{2}}}}{4}\]
D) \[\frac{b({{a}^{2}}+{{b}^{2}})}{4a}\]
Correct Answer: D
Solution :
Given curve \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] and point P \[\left( \frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}} \right)\] Slope of tangent at P is\[-\text{ }b/a\]. Slope of normal at P is a/b. Equation of tangent on the curve \[bx+ay=\sqrt{2}ab\] ... (i) Equation of normal on the tangent of the curve \[-ax+by=\frac{({{b}^{2}}-{{a}^{2}})}{\sqrt{2}}\] ...(ii) Equation of x-axis \[y=0\] ... (iii) Solving these equations in successive manner, we get three points of triangle \[\left( \frac{{{a}^{2}}-{{b}^{2}}}{a\sqrt{2}},0 \right)(\sqrt{2}.a,0)\] and \[\left( \frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}} \right)\] Now, area of triangle \[=\frac{1}{2}\left| \begin{matrix} \frac{{{a}^{2}}-{{b}^{2}}}{a\sqrt{2}} & 0 & 1 \\ \sqrt{2}.a & 0 & 1 \\ a/\sqrt{2} & b/\sqrt{2} & 1 \\ \end{matrix} \right|\] \[=\frac{-b}{2\sqrt{2}}\left\{ \frac{{{a}^{2}}-{{b}^{2}}}{a\sqrt{2}}-\sqrt{2}.a \right\}=\frac{-b}{4a}(-{{a}^{2}}-{{b}^{2}})\] \[=\frac{b({{a}^{2}}+{{b}^{2}})}{4a}\]
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