A) \[a\text{ }\frac{{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{3/2}}}{{{h}^{2}}+{{k}^{2}}}\]
B) \[a\text{ }\frac{{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{1/2}}}{{{h}^{2}}+{{k}^{2}}}\]
C) \[\frac{{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{3/2}}}{{{h}^{2}}+{{k}^{2}}}\]
D) \[\frac{{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{1/2}}}{{{h}^{2}}+{{k}^{2}}}\]
Correct Answer: A
Solution :
Equation of chord of contact AB is \[xh+yk={{a}^{2}}\] .....(i) \[OM=\]length of perpendicular from O(0, 0) on line (i) \[=\frac{{{a}^{2}}}{\sqrt{{{h}^{2}}+{{k}^{2}}}}\] \[\therefore \] \[AB=2AM=2\sqrt{O{{A}^{2}}-O{{M}^{2}}}=\frac{2a\sqrt{{{h}^{2}}+{{k}^{2}}-{{a}^{2}}}}{\sqrt{{{h}^{2}}+{{k}^{2}}}}\] Also \[PM=\]length of perpendicular from \[P(h,\ k)\]to the line (i) is \[\frac{{{h}^{2}}+{{k}^{2}}-{{a}^{2}}}{\sqrt{{{h}^{2}}+{{k}^{2}}}}\] Therefore, the required area of triangle PAB \[=\frac{1}{2}.\ AB\ .\ PM=\frac{a{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{3/2}}}{{{h}^{2}}+{{k}^{2}}}\].You need to login to perform this action.
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