A) \[4a\cos \text{ec}\alpha \cdot \cot \alpha \]
B) \[4a\tan \alpha \cdot \sec \alpha \]
C) \[4a\cos \alpha \cdot \cot \alpha \]
D) \[4a\sin \alpha \cdot \tan \alpha \]
Correct Answer: B
Solution :
Let \[A\] be the vertex and \[AP\] be a chord of \[{{x}^{2}}=4ay\] such that slope of \[AP\] is\[\tan \alpha \]. Let the coordinates of \[P\] be\[(2at,\,\,a{{t}^{2}})\]then, Slope of\[AP=\frac{a{{t}^{2}}}{2at}=\frac{t}{2}\] \[\Rightarrow \] \[\tan \alpha =\frac{t}{2}\] \[\Rightarrow \] \[t=2\tan \alpha \] Now, \[AP=\sqrt{{{(2at-0)}^{2}}+{{(a{{t}^{2}}-0)}^{2}}}\] \[=at\sqrt{4+{{t}^{2}}}=2a\tan \alpha \sqrt{4+4{{\tan }^{2}}\alpha }\] \[=4a\tan \alpha \cdot \sec \alpha \]
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