A) 9
B) \[-\,5\]
C) 7
D) \[-\,7\]
Correct Answer: A
Solution :
[a] \[{{y}^{2}}=\alpha {{x}^{3}}-\beta \] or \[\frac{dy}{dx}=\frac{3\alpha {{x}^{2}}}{2y}\] Therefore, slope of the normal at (2, 3) is \[{{\left( -\frac{dx}{dy} \right)}_{(2,3)}}=-\frac{2\times 3}{3\alpha {{(2)}^{2}}}=-\frac{1}{2\alpha }=-\frac{1}{4}\] Or \[\alpha =2\] Also, (2, 3) lies on the curve. Therefore, \[9=8\alpha -\beta \] or \[\beta =16-9=7\] or \[\alpha +\beta =9.\]
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