Find p and q, if the equation 2x²+8xy+qx+2y-15=0 represents a pair of parallel lines
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Given Find p and q, if the equation 2x²+ 8xy + py^2 + qx + 2y - 15 = 0 represents a pair of parallel lines
- Now the general equation is given by ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0
- In this equation angle between lines we have
- So tan theta = 2 √h^2- ab / a + b
- So for parallel lines we have h^2 = ab
- Now from the equation 2h = 8 or h = 4
- Also a = 2
- And b = p
- Substituting the values in h^2 = ab we get
- 16 = 2 p
- Or p = 8
- Now in the given curve (pair of lines) we have the determinant as
- So a h g
- So h b f = 0
- g f c
- 2 4 q/2
- 4 8 1 = 0
- So q/2 1 - 15
- So 2(- 120 – 1) – 4(- 60 – q/2) + q/2 (4 – 8q / 2) = 0
- 242 + 240 + 4q – 2q^2
- 2q^2 – 4q + 2 = 0
- 2(q^2 – 2q + 1) = 0
- Or q^2 – q – q + 1 = 0
- Or q(q – 1) – 1(q – 1) = 0
- Or q = 1
- Therefore we get p = 8 and q = 1
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