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Find equations of both tangent
發問:
2)Find the equations of both the tangent lines to the ellipse x^2 + 4y^2= 36 that pass through the point (2,3) Please step by step Thanks a lot
Answer: y = 3 and 3x + 8y - 30 = 0 Let the equation be y - 3 = m(x - 2) => y = m(x - 2) + 3 Sub. y = m(x - 2) + 3 into x^2 + 4y^2 = 36 x^2 + 4[m(x - 2) + 3]^2 = 36 x^2 + 4m^2 (x - 2)^2 + 24m(x - 2) + 36 = 36 x^2 + 4m^2 x^2 - 16m^2 x + 16m^2 + 24mx - 48m + 36 = 36 (1 + 4m^2) x^2 + (24m - 16m^2) x + 16m^2 - 48m = 0 since the tangents cut the ellipse at one point, (24m - 16m^2)^2 - 4(1 + 4m^2)(16m^2 - 48m) = 0 (12m - 8m^2)^2 - (1 + 4m^2)(16m^2 - 48m) = 0 (64m^4 - 192m^3 + 144m^2) - (64m^4 - 192m^3 + 16m^2 - 48m) = 0 128m^2 + 48m = 0 m = 0 or m = -3/8 thus, the equation of tangent are y - 3 = 0 and y - 3 = (-3/8)(x - 2) i.e. y = 3 and 3x + 8y - 30 = 0 2013-05-21 11:16:36 補充： Methos 2 Let the intersection point be (a, b) differentiate both sides of x^2 + 4y^2 = 36 w.r.t. x 2x + 8y(dy/dx) = 0 => dy/dx = -x/(4y) 2013-05-21 11:16:57 補充： Solve -a/(4b) = (b - 3)/(a - 2) ------ (i) and a^2 + 4b^2 = 36 ------ (ii) from (i), -a^2 + 2a = 4b^2 - 12b => a^2 + 4b^2 - 2a - 12b = 0 ------ (iii) (iii) - (ii), a = 18 - 6b ------ (iv) Sub. (iv) into (ii), (18 - 6b)^2 + 4b^2 = 36 10b^2 - 54b + 72 = 0 b = 3 or 12/5 2013-05-21 11:17:56 補充： when b = 3, a = 0, dy/dx = 0, the equation of tangent is y = 3 when b = 12/5, a = 18/5, dy/dx = -3/8, the equation of tangent is y - 12/5 = (-3/8)(x - 18/5) => 3x + 8y - 30 = 0
其他解答:
Find equations of both tangent
發問:
2)Find the equations of both the tangent lines to the ellipse x^2 + 4y^2= 36 that pass through the point (2,3) Please step by step Thanks a lot
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最佳解答:Answer: y = 3 and 3x + 8y - 30 = 0 Let the equation be y - 3 = m(x - 2) => y = m(x - 2) + 3 Sub. y = m(x - 2) + 3 into x^2 + 4y^2 = 36 x^2 + 4[m(x - 2) + 3]^2 = 36 x^2 + 4m^2 (x - 2)^2 + 24m(x - 2) + 36 = 36 x^2 + 4m^2 x^2 - 16m^2 x + 16m^2 + 24mx - 48m + 36 = 36 (1 + 4m^2) x^2 + (24m - 16m^2) x + 16m^2 - 48m = 0 since the tangents cut the ellipse at one point, (24m - 16m^2)^2 - 4(1 + 4m^2)(16m^2 - 48m) = 0 (12m - 8m^2)^2 - (1 + 4m^2)(16m^2 - 48m) = 0 (64m^4 - 192m^3 + 144m^2) - (64m^4 - 192m^3 + 16m^2 - 48m) = 0 128m^2 + 48m = 0 m = 0 or m = -3/8 thus, the equation of tangent are y - 3 = 0 and y - 3 = (-3/8)(x - 2) i.e. y = 3 and 3x + 8y - 30 = 0 2013-05-21 11:16:36 補充： Methos 2 Let the intersection point be (a, b) differentiate both sides of x^2 + 4y^2 = 36 w.r.t. x 2x + 8y(dy/dx) = 0 => dy/dx = -x/(4y) 2013-05-21 11:16:57 補充： Solve -a/(4b) = (b - 3)/(a - 2) ------ (i) and a^2 + 4b^2 = 36 ------ (ii) from (i), -a^2 + 2a = 4b^2 - 12b => a^2 + 4b^2 - 2a - 12b = 0 ------ (iii) (iii) - (ii), a = 18 - 6b ------ (iv) Sub. (iv) into (ii), (18 - 6b)^2 + 4b^2 = 36 10b^2 - 54b + 72 = 0 b = 3 or 12/5 2013-05-21 11:17:56 補充： when b = 3, a = 0, dy/dx = 0, the equation of tangent is y = 3 when b = 12/5, a = 18/5, dy/dx = -3/8, the equation of tangent is y - 12/5 = (-3/8)(x - 18/5) => 3x + 8y - 30 = 0
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