A point P moves such that sum of the slopes of the normals drawn from it to the hyperbola xy = 16 is equal to the sum of ordinates of feet of normals. Which of the following is/are True

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Product of abscissa of feet of normals is -1

Prodcut of abscissa of feet of normals is -256

Locus of P is $x^{2} = 12 y$

Locus of P is $x^{2} = 16 y$

answer is B, C.

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Detailed Solution

$\begin{array}{l} x y = 16 \Rightarrow Q (4 t, \frac{4}{t}) \\ Normal at Q is y - \frac{4}{t} = t^{2} (x - 4 t) \\ It passes through P (h, k) \Rightarrow \Rightarrow \sum t_{1} = \frac{h}{4}, \sum t_{1} t_{2} = 0, \sum t_{1} t_{2} t_{3} = - \frac{k}{4} \\ \Rightarrow y_{1} + y_{2} + y_{3} + y_{4} = k \\ 16 (t_{1}^{2} + t_{2}^{2} + t_{3}^{2} + t_{4}^{2}) = h^{2} \\ But given sum of slopes of normal = sum of feet of normals \\ \frac{h^{2}}{16} = k \Rightarrow x^{2} 16 y \end{array}$