Introduction:
Two points define a straight line. Three points define a circle. How many points should be needed to define a conic? Five! Why? |
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The general equation has 6 coefficients
: Divide the equation through-out by the constant A. We get: So we need five points to find all five coefficients: p, q, r, s, t. Solving 5 equations with 5 unknowns is not an easy task……. |
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Find the equation of the conic passes through the
points :
P(0, -4), Q(1, -6), R(3, 2), S(4, 1), T(2, -1) |
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PQ : 2x +y + 4
= 0 RS : x + y – 5
= 0 PS : 5x – 4y –
16 = 0 QR : 4x – y –
10 = 0 |
Use Two points form to find the straight lines. |
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PQ´RS : (2x + y + 4)(x + y –
5) = 0 PS´QR : (5x – 4y – 16)(4x – y
– 10) = 0 |
Two pairs of straight lines, each passing through P, Q, R, S. |
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System of conics passing
through P, Q, R, S: PQ´RS+ k PS´QR : (2x + y + 4)(x + y – 5) +
k(5x – 4y – 16)(4x – y – 10)= 0
(1) |
Forming family of conics through 4 points. |
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Since T(2, -1) is on (1),
substitution will get: -28 + 2k = 0 \ k = 14 ….. (2) Subst. (2) in (1), we get: (2x + y + 4)(x + y – 5) + 14(5x – 4y – 16)(4x – y – 10)= 0 |
Use the last point to find k. |
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The required conic : |
Simplify |
