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1條 amaths circle 急
given a circle C : x^2+y^2 = 1 and a point P (h,k) outside C (a) If L:y=mx+c is tangent to C , show that c^2=m^2+1 (b) If L passes through the point P Show that (h^2-1)m^2 -2hkm +k^2 -1 =0 (c) hence fing the equation of the locus of P such that two tangents from P make an angle of 45degree 更新: 想問下點解e兩步點變出來 (m - m')^2 = (1 + mm')^2 (m + m')^2 - 4mm' = (1 + mm')^2 更新 2: 我明啦 十萬唔該....
最佳解答:
a. C: x^2 + y^2 = 1 ... (1) L: y = mx + c ... (2) Put (2) into (1): x^2 + (mx + c)^2 = 1 (m^2 + 1)x^2 + 2mcx + (c^2 - 1) = 0 discriminant = 0, for L being a tangent to C (2mc)^2 - 4(m^2 + 1)(c^2 - 1) = 0 4m^2c^2 - 4m^2c^2 + 4m^2 - 4c^2 + 4 = 0 c^2 = m^2 + 1 b. Since L passes through P(h, k) So, k = mh + c c = k - mh ... (3) Put (3) into the result of a (k - mh)^2 = m^2 + 1 k^2 - 2hkm + m^2h^2 = m^2 + 1 (h^2 - 1)m^2 - 2hkm + (k^2 - 1) = 0 c. Since the two tangents satisfy the solution in (b). And the two tangents make an angle of 45* So, │(m - m') / (1 + mm')│ = tan45* (m - m')^2 = (1 + mm')^2 (m + m')^2 - 4mm' = (1 + mm')^2 By sum of roots and product of roots, (2hk/(h^2 - 1))^2 - 4(k^2 - 1)/(h^2 - 1) = [1 + (k^2 - 1)/(h^2 - 1)]^2 4h^2k^2 - 4(k^2 - 1)(h^2 - 1) = [(h^2 - 1) + (k^2 - 1)]^2 4h^2k^2 - 4h^2k^2 + 4k^2 + 4h^2 - 4 = (h^2 + k^2)^2 - 4(h^2 + k^2) + 4 8h^2 + 8k^2 - 8 = h^4 + 2h^2k^2 + k^4 Equation of locus: x^4 + 2x^2y^2 + y^4 - 8x^2 - 8y^2 + 8 = 0
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1條 amaths circle 急
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發問:given a circle C : x^2+y^2 = 1 and a point P (h,k) outside C (a) If L:y=mx+c is tangent to C , show that c^2=m^2+1 (b) If L passes through the point P Show that (h^2-1)m^2 -2hkm +k^2 -1 =0 (c) hence fing the equation of the locus of P such that two tangents from P make an angle of 45degree 更新: 想問下點解e兩步點變出來 (m - m')^2 = (1 + mm')^2 (m + m')^2 - 4mm' = (1 + mm')^2 更新 2: 我明啦 十萬唔該....
最佳解答:
a. C: x^2 + y^2 = 1 ... (1) L: y = mx + c ... (2) Put (2) into (1): x^2 + (mx + c)^2 = 1 (m^2 + 1)x^2 + 2mcx + (c^2 - 1) = 0 discriminant = 0, for L being a tangent to C (2mc)^2 - 4(m^2 + 1)(c^2 - 1) = 0 4m^2c^2 - 4m^2c^2 + 4m^2 - 4c^2 + 4 = 0 c^2 = m^2 + 1 b. Since L passes through P(h, k) So, k = mh + c c = k - mh ... (3) Put (3) into the result of a (k - mh)^2 = m^2 + 1 k^2 - 2hkm + m^2h^2 = m^2 + 1 (h^2 - 1)m^2 - 2hkm + (k^2 - 1) = 0 c. Since the two tangents satisfy the solution in (b). And the two tangents make an angle of 45* So, │(m - m') / (1 + mm')│ = tan45* (m - m')^2 = (1 + mm')^2 (m + m')^2 - 4mm' = (1 + mm')^2 By sum of roots and product of roots, (2hk/(h^2 - 1))^2 - 4(k^2 - 1)/(h^2 - 1) = [1 + (k^2 - 1)/(h^2 - 1)]^2 4h^2k^2 - 4(k^2 - 1)(h^2 - 1) = [(h^2 - 1) + (k^2 - 1)]^2 4h^2k^2 - 4h^2k^2 + 4k^2 + 4h^2 - 4 = (h^2 + k^2)^2 - 4(h^2 + k^2) + 4 8h^2 + 8k^2 - 8 = h^4 + 2h^2k^2 + k^4 Equation of locus: x^4 + 2x^2y^2 + y^4 - 8x^2 - 8y^2 + 8 = 0
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