Question

If the equation of the tangent drawn at \((h, k)\) to the hyperbola \(\frac{(x - 1)^2}{1} - \frac{(y - 2)^2}{2} = 1\) is \(x = 2\), then \(h + k =\)

• A. \(0\)
• B. \(4\)
• C. \(-4\)
• D. \(1\)

Hint:

For a vertical tangent to a hyperbola, the term with \(y\) must cancel out in the tangent equation.

Answer 1

The Correct Option is B

The equation of the hyperbola is \[ \frac{(x - 1)^2}{1} - \frac{(y - 2)^2}{2} = 1 \] The equation of the tangent to a hyperbola at point \((h, k)\) is: \[ \frac{(x - 1)(h - 1)}{1} - \frac{(y - 2)(k - 2)}{2} = 1 \] Given that this tangent is vertical (i.e., \(x = 2\)), it must be independent of \(y\). That means the coefficient of \((y - 2)\) must be 0: \[ \frac{(x - 1)(h - 1)}{1} - 0 = 1 \Rightarrow (x - 1)(h - 1) = 1 \] Since the line is \(x = 2\), substitute \(x = 2\): \[ (2 - 1)(h - 1) = 1 \Rightarrow (h - 1) = 1 \Rightarrow h = 2 \] Now use the tangent equation again: \[ \frac{(x - 1)(2 - 1)}{1} - \frac{(y - 2)(k - 2)}{2} = 1 \Rightarrow (x - 1) - \frac{(y - 2)(k - 2)}{2} = 1 \] Since \(x = 2\): \[ (2 - 1) - \frac{(y - 2)(k - 2)}{2} = 1 \Rightarrow 1 - \frac{(y - 2)(k - 2)}{2} = 1 \Rightarrow \frac{(y - 2)(k - 2)}{2} = 0 \] So \((y - 2)(k - 2) = 0\). This gives either \(y = 2\) or \(k = 2\). But this must hold for **all** \(y\) on the line, so the only valid constant is \(k = 2\). Thus, \(h = 2\), \(k = 2\), so: \[ h + k = 2 + 2 = 4 \]