Question

# Match the column EquationName of the curve1)x2−2x−y−3=0P) Circle2)x2+3xy+2y2−x−4y−6=0Q) Parabola3)x2+y2−20=0R) Ellipse4)7x2+7y2+2xy+10x−10y+7=0S) Hyperbola5)6x2−xy−y2−23x+4y+15=0T) Pair of straight lines1 - T, 2 - Q, 3 - P, 4 - R, 5 - S  1 - Q, 2 - T, 3 - P, 4 - R, 5 - S 1 - Q, 2 - T, 3 - P, 4 - S, 5 - R 1 - Q, 2 - S, 3 - P, 4 - T, 5 - R

Solution

## The correct option is B 1 - Q, 2 - T, 3 - P, 4 - R, 5 - S We have to decide the curve represented by the equations given. All of them are second degree curves. The systematic way of finding the conic is as follows. Let the equation of the curve be ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 and △ = abc + 2fgh − af2 − bg2 − ch2 1) Find △ 2) If △ = 0, it represents a pair of straight lines 3) If △ ≠ 0, it's a conic No we have to decide which conic for that we will compare h2 and ab. 4) h2 = ab ⇒ parabola 5) h2 < ab ⇒ ellipse or circle For circle, a = b and h = 0 otherwise it will be an ellipse 6) h2 > ab ⇒ hyperbola We will go through each equation and decide P) x2 − 2x − y − 3 = 0 a = 1,b = 0,h = 0,g = −1,f = −12,c = −3 △ = 0+0−1 × (−12)2 −0 − 0=−14 △ ≠ 0 h2 = 0 ab = 0 △ = 0 and h2 = ab ⇒ parabola Q) x2 + 3xy + 2y2 − x − 4y − 6 = 0 a = 1,b = 2,b=32,g=−12,f=−2,c=−6 △ = 1 × 2 × −6+2 ×−2 × −12 × 32 −1 × (−2)2 −2 × (−12)2 − (−6) × 94 △ = 0 △ = 0 ⇒ pair of straight lines R) x2 + y2 − 20 = 0 If a=b and h=0 ,it represents a circle Here a =b=1 and h=0 ⇒ circle S) 7x2 + 7y2 + 2xy + 10x − 10y + 7 = 0 a=7,b=7,h=1,g=5,f=−5,c=7 △ = 7 × 7 × 7 + 2 × −5 × 5 × 1−7 × (−5)2 −7 × 52−7 × 12 △ ≠ 0 h2 = 1,ab = 44 △ ≠ 0,b2 < ab ⇒ Ellipse T) 6x2 − xy − y2 − 23x + 4y + 15 = 0 a=6,b=−1,h=−12,g=−232,f=2,c=15 △ ≠ 0 h2 = (−12)2 = 14,ab=−6 h2 > ab ⇒ Hyperbola

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