Question

The curve represented by \(\frac{x^2}{12-\alpha} + \frac{y^2}{\alpha-10} = 1\) is

• A. a hyperbola for some values of \(\alpha\) in \((10,12)\)
• B. an ellipse for all values of \(\alpha\) in \((10,12)\)
• C. a circle for some value of \(\alpha\) in \((10,12)\)
• D. a hyperbola for all values of \(\alpha\) in \((10,12)\)

Hint:

An equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) represents a circle if \(a^2 = b^2\). Always check if the denominators can be equal within the given domain.

Answer 1

The Correct Option is C

Step 1: Analyze the Equation:
The equation is \(\frac{x^2}{A} + \frac{y^2}{B} = 1\) where \(A = 12-\alpha\) and \(B = \alpha-10\). For \(\alpha \in (10, 12)\): \(12 - \alpha>0\) (Positive) \(\alpha - 10>0\) (Positive) Since both denominators are positive, the curve represents an ellipse generally.
Step 2: Check for Circle Condition:
For the curve to be a circle, the coefficients of \(x^2\) and \(y^2\) must be equal, i.e., \(A = B\). \[ 12 - \alpha = \alpha - 10 \] \[ 2\alpha = 22 \implies \alpha = 11 \] The value \(\alpha = 11\) lies in the interval \((10, 12)\). Thus, for \(\alpha = 11\), the curve is a circle.
Step 3: Evaluate Options:
(A) It is never a hyperbola in this interval because \(A\) and \(B\) are both positive. (B) It is an ellipse for \(\alpha \neq 11\), but specifically a circle at \(\alpha = 11\). Usually "ellipse" includes the special case of a circle, but option (C) is a more specific and distinct existence claim. (C) It becomes a circle for \(\alpha = 11\), which is in the range. This is true.