Correct Answer - Option 2 : 2
Concept:
Let second-degree equation be,
ax2 + by2 + 2hxy + 2gx + 2fy + c = 0
Discriminant is calculated as:
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 or pair of perpendicular straight lines
- If h2 – ab = 0, it represents a parallel lines.
- If h2 – ab < 0, it represents non-real lines.
Calculation:
Given: Second degree equation, 3x2 + 7xy + 2y2 + 5x + 5y + k = 0
Compare with second-degree equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
So, a = 3, h = 3.5, b = 2, g = 2.5, f = 2.5, c = k
The given equation represents the pair of straight lines so;
Δ = 0
abc + 2fgh - af2 - bg2 - ch2 = 0
3 x 2 x k + 2 x 2.5 x 2.5 x 3.5 - 3 x (2.5)2 - 2 x (2.5)2 - k x (3.5)2 = 0
6k + 43.75 - 18.75 - 12.5 - 12.25k = 0
6.25k = 12.5
k = 2