Question

If \( S = \frac{x^2}{k - 7} - \frac{y^2}{11 - k} = 0, k \in \mathbb{R}, k \neq 7,11 \), then which one of the following statements is incorrect?

• A. \( S = 0 \) represents a circle with radius \( \sqrt{2} \), when \( k = 9 \)
• B. \( S = 0 \) represents an ellipse with eccentricity \( \frac{\sqrt{2}}{3} \), when \( k = 10 \)
• C. \( S = 0 \) represents a hyperbola with eccentricity \( \frac{\sqrt{6}}{5} \), when \( k = 12 \)
• D. \( S = 0 \) represents a hyperbola with eccentricity \( \frac{\sqrt{3}}{2} \), when \( k = 13 \)

Hint:

For conic sections, identify the type based on the sign and values of the denominators. The eccentricity can be calculated based on the coefficients.

Answer 1

The Correct Option is B

We are given the equation of a conic: \[ \frac{x^2}{k - 7} - \frac{y^2}{11 - k} = 0 \] This represents a conic whose type depends on the value of \( k \). The nature of the conic (circle, ellipse, hyperbola) depends on the signs of the terms in the equation. Step 1: Conditions for a circle, ellipse, or hyperbola - Circle: \( \frac{x^2}{r^2} + \frac{y^2}{r^2} = 0 \) — when \( k = 9 \) - Ellipse: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0 \) — when \( k = 10 \) - Hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0 \) — when \( k = 12 \) Step 2: Check the values for each case - For \( k = 9 \), the equation represents a circle with radius \( \sqrt{2} \), which is correct. - For \( k = 10 \), the equation represents an ellipse, but the eccentricity does not match the given value \( \frac{\sqrt{2}}{3} \), so this statement is incorrect. - For \( k = 12 \), the equation represents a hyperbola with eccentricity \( \frac{\sqrt{6}}{5} \), which is correct. - For \( k = 13 \), the equation represents a hyperbola with eccentricity \( \frac{\sqrt{3}}{2} \), which is also correct. Thus, the correct answer is option (2), which is incorrect.

Answer 2

Step 1: Understand the given equation
We are given:
S = (x²)/(k − 7) − (y²)/(11 − k) = 0, where k ∈ ℝ and k ≠ 7, 11
This is a second-degree equation in x and y and resembles the standard form of a conic section:
(x²/a²) − (y²/b²) = 1 (hyperbola) or (y²/b²) − (x²/a²) = 1 (also hyperbola), depending on signs

However, in our case the RHS is 0, not 1. But:
(x²)/(A) − (y²)/(B) = 0 can be rewritten as:
(x²)/(A) = (y²)/(B) ⇒ Bx² = Ay²
This is the general form of a pair of straight lines or a degenerate conic unless modified further.

Step 2: Multiply both sides by denominator
(x²)/(k−7) − (y²)/(11−k) = 0 ⇒
[(11 − k)x² − (k − 7)y²]/[(k−7)(11−k)] = 0 ⇒ numerator must be zero
So we get:
(11 − k)x² − (k − 7)y² = 0

This is a homogeneous second-degree equation in x and y. It represents a pair of straight lines passing through the origin (degenerate conic), **not an ellipse or hyperbola**.

Step 3: Consider the nature of the conic based on coefficients
Equation: (x²)/(k − 7) − (y²)/(11 − k) = 0
Let’s define a = √|k − 7|, b = √|11 − k|
If both (k−7) and (11−k) are positive ⇒ (k ∈ (7, 11)) ⇒ Equation becomes of form (x²/a²) − (y²/b²) = 0 ⇒ hyperbola
If both are negative ⇒ k ∈ (−∞, 7) ∪ (11, ∞), same result as above since signs match
But if they have opposite signs ⇒ (x²/a²) + (y²/b²) = 0 ⇒ **no real curve unless zero**, or imaginary ellipse

Step 4: Analyze the specific claim
Check for k = 10:
Then (x²)/(3) − (y²)/(1) = 0 ⇒ x²/3 = y² ⇒ cross-multiplied: x² = 3y² ⇒ 3y² − x² = 0
This is a pair of straight lines through the origin, not an ellipse

Also, the given statement says:
“S = 0 represents an ellipse with eccentricity √2/3 when k = 10”
This is incorrect because:
(1) The equation S = 0 is not of ellipse form when k = 10
(2) It reduces to a pair of lines and not a closed conic

Final Conclusion:
The statement is incorrect because for k = 10, the equation does not represent an ellipse; instead, it represents a degenerate conic (pair of straight lines)