Question

The conic x² + xy + 2y² + x + y = 1 is
1. an ellipse
2. a hyperbola
3. a parabola
4. a pair of straight lines

Answer 1

Correct Answer - Option 1 : an ellipse

Concept:

Let second-degree equation be,

ax2 + by2 + 2hxy + 2gx + 2fy + c = 0

Discriminant is:

Δ = \(\left| {\begin{array}{*{20}{c}} a&h&{\rm{g}}\\ h&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right|\)

Δ = abc + 2fgh - af2 - bg2 - ch2

Case: 1

If discriminant (Δ) of this equation doesn’t equal to zero (Δ ≠ 0)

• If h2 – ab > 0, it represents a hyperbola and a rectangular hyperbola (a + b = 0).
• If h2 – ab = 0, it represents a parabola.
• If h2 – ab < 0, it represents an ellipse. (a ≠ b)
• If h2 – ab < 0, it represents a circle. (a = b)

Case: 2

If discriminant (Δ) of this equation equal to zero (Δ = 0)

• If h2 – ab > 0, it represents two distinct real lines.
• If h2 – ab = 0, it represents a parallel lines.
• If h2 – ab < 0, it represents non-real lines.

 

Calculation:

Given: Second degree equation, x2 + xy + y2 + x + y = 1 

Compare with second-degree equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

So, a = 1, b = 2, h = 0.5, g = 0.5, f = 0.5 and c = -1

Now, Δ = abc + 2fgh - af2 - bg2 - ch2

Δ = 1 × 2 × (-1) + 2 x 0.5 x 0.5 x 0.5 – 1 × (0.5)2 – 2 × (0.5)2 - (-1) x (0.5)²

= -1 – 0.25 - 0.25 - 0.5 + 0.25

So, Δ = -1.75

Now, h2 – ab = (0.5)2 – 1 × 2 = -1.75

So, h2 – ab < 0 and a  ≠ b

It represents a ellipse.