The parameter equation ellipse 2x square + 3Y square = 6 has the shortest distance to the line 2x + 3y-6 = 0 In the previous steps, I will, d = | 2 √ 3cosa + 3 √ 2sina-6 | / √ (2 ^ 2 + 3 ^ 2) And then I won't Teach me the method Actually, I don't know the auxiliary angle formula! Is it √ A & # 178; + B & # 178; then it must be sin (not COS)?

The parameter equation ellipse 2x square + 3Y square = 6 has the shortest distance to the line 2x + 3y-6 = 0 In the previous steps, I will, d = | 2 √ 3cosa + 3 √ 2sina-6 | / √ (2 ^ 2 + 3 ^ 2) And then I won't Teach me the method Actually, I don't know the auxiliary angle formula! Is it √ A & # 178; + B & # 178; then it must be sin (not COS)?

D = | 2 √ 3cosa + 3 √ 2sina-6 | / √ (2 ^ 2 + 3 ^ 2) denominator is constant, because finding the minimum value, so regardless of it (add it at last) | 2 √ 3cosa + 3 √ 2sina-6 | auxiliary angle formula = | √ [(2 √ 3) ^ 2 + (3 √ 2) ^ 2] sin (a + b) - 6 | = | √ 30sin (a + b) - 6 | the minimum value is √ 30-6, and the shortest distance is (...)

Given that P is a point on an ellipse 2x2 + 3y2 = 6, then the shortest distance from P to a focal point of the ellipse is______ ．

∵ ellipse 2x2 + 3y2 = 6, ∵ x23 + Y22 = 1, ∵ a = 3, B = 2, C = 1, ∵ p the shortest distance to a focus of the ellipse is: a-c = 3-1

Find a point on the square of ellipse x + 4Y square = 4 to make it the shortest distance to the line 2x + 3y-6 = 0

Let a straight line 2x + 3Y + B = 0, and then tangent to the ellipse, and then get b = several. Then calculate the distance between this straight line and the original straight line

Ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ = 1, (a > b > 0) eccentricity is √ 3 / 2, a + B = 3,
(1) Solving elliptic equation
(2) The slope of BP is K and that of Mn is m, which proves that 2m-k is a fixed value

As shown in the figure, a, B, D are the vertex of ellipse C, P is any point except the vertex of ellipse C, the straight line DP intersects X axis at point n, the straight line ad intersects BP at point m, let the slope of BP be K, the slope of Mn be m, and prove that 2m-k is the fixed value. I'm glad to answer for you, and wish you progress in your study! If you don't understand me, please choose it as a satisfactory answer, thank you!

High school mathematics problem ellipse over focus (- C, 0) and ellipse intersection ratio 2:1 to find eccentricity
In five ways

Method 1. Polar coordinate method
Method 2, plane geometry method
Method 3, definition method
Method 4, the second definition method
Method 5. Parametric equation method
Method 6, the method of transformation
Methods 7. Observation
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I've been studying conic for 35 years, and I only do this one thing in my life. You don't know what to do with this problem

It is known that the eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a ＞ B ＞ 0) is √ 2 / 2, and the standard equation of ellipse C is obtained by passing through point (1 / 2, √ 14 / 4)

(theorem) if we know that curve C1 equation is x2 − Y28 = 1 (x ≥ 0, y ≥ 0), circle C2 equation is (x-3) 2 + y2 = 1, slope is K (k > 0), line L is tangent to circle C2, tangent point is a, line L intersects curve C1 at point B, | ab | = 3, then the slope of line AB is ()
A. 1B. 12C. 33D. 3

The center of circle C2 is the right focus of hyperbola ∵ ab | = 3, the radius of circle C2 is 1 ∵ BC2 | = 2, let the coordinates of B be (x, y), (x > 0) ∵ the right collimator of hyperbola is x = 13 ∵ 2x − 13 = 3 ∵ x = 1 ∵ B (1, 0). Let the equation of AB be y = K (x-1), that is, kx-y-k = 0