Question

If the normal drawn to the hyperbola \( xy = 16 \) at (8, 2) meets the hyperbola again at a point \((\alpha, \beta)\), then \( |\beta| + \frac{1}{|\alpha|} = \)

• A. 40
• B. 34
• C. 28
• D. 54

Hint:

For the hyperbola \( xy = c^2 \), the normal at \((x_1, y_1)\): \( x x_1 - y y_1 = x_1^2 - y_1^2 \). Solve with the hyperbola equation to find intersection points.

Answer 1

The Correct Option is B

The hyperbola \( xy = 16 \) is a rectangular hyperbola (\( c^2 = 16 \), \( c = 4 \)). The normal at \((x_1, y_1)\) is: \[ x x_1 - y y_1 = x_1^2 - y_1^2 \] At \((8, 2)\): \[ 8x - 2y = 8^2 - 2^2 = 64 - 4 = 60 \] \[ 4x - y = 30 \] The point \((\alpha, \beta)\) lies on the hyperbola (\( \alpha \beta = 16 \)) and the normal: \[ 4\alpha - \beta = 30 \implies \beta = 4\alpha - 30 \] \[ \alpha (4\alpha - 30) = 16 \implies 4\alpha^2 - 30\alpha - 16 = 0 \implies 2\alpha^2 - 15\alpha - 8 = 0 \] Solve: \[ \alpha = \frac{15 \pm \sqrt{225 + 64}}{4} = \frac{15 \pm \sqrt{289}}{4} = \frac{15 \pm 17}{4} \] \[ \alpha = \frac{32}{4} = 8 \quad \text{or} \quad \alpha = \frac{-2}{4} = -\frac{1}{2} \] Exclude \(\alpha = 8\) (original point). Use \(\alpha = -\frac{1}{2}\): \[ \beta = \frac{16}{-\frac{1}{2}} = -32 \] \[ |\beta| + \frac{1}{|\alpha|} = |-32| + \frac{1}{\left|-\frac{1}{2}\right|} = 32 + 2 = 34 \] Option (2) is correct. Options (1), (3), and (4) do not match.