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question:Given the ellipse frac {x^{2}}{a^{2}}+ frac {y^{2}}{b^{2}}=1(a>b>0), a circle is drawn with its center at the right focus F_2 of the ellipse. This circle passes through the center of the ellipse and intersects the ellipse at points M and N. If the line F_1M passing through the left focus F_1 of the ellipse is a tangent to the circle F_2, find the eccentricity of the ellipse.
answer:From the given information, we have: - |MF_2|=|OF_2|=c - |MF_1|+|MF_2|=2a - |F_1F_2|=2c In the right triangle MF_1F_2, we have: - |MF_1|^2+|MF_2|^2=|F_1F_2|^2 Substituting the given values, we get: - (2a-c)^2+c^2=4c^2 Simplifying the equation, we get: - 2a^2-2ac-c^2=0 This equation can be rewritten in terms of eccentricity e as: - e^2+2e-2=0 Solving this equation, we get e=sqrt{3}-1 or e=-sqrt{3}-1 (which is discarded as eccentricity cannot be negative). Therefore, the answer is: boxed{e=sqrt{3}-1}. This problem primarily tests the ability to analyze and solve problems involving lines and conic sections. The solution is derived based on the given information and the definition of an ellipse, using the Pythagorean theorem to establish the equation and find the eccentricity.
question:An edge of a square can be expressed as 4x15 meters or as 203x meters. What is its area in square meters?
answer:To find the edge of the square, we first set the two expressions for the edge equal to each other: [ 4x - 15 = 20 - 3x. ] Solving for x, we add 3x to both sides and add 15 to both sides to get: [ 4x + 3x = 20 + 15 7x = 35. ] Dividing both sides by 7 gives us: [ x = frac{35}{7} x = 5. ] Now, substituting x = 5 into either expression for the edge of the square, let's use 4x - 15: [ 4(5) - 15 = 20 - 15 = 5 text{ meters}. ] The side length of the square is 5 meters. To find the area of the square, we square this length: [ text{Area} = (5text{ m})^2 = 25 text{ square meters}. ] Therefore, the area of the square is boxed{25} square meters.
question:John plays paintball 3 times a month on average, but in peak season he plays twice more frequently. Each time he plays, he buys 3 boxes of paintballs. The cost per box varies based on the quantity purchased: 1 to 5 boxes: 25 per box 6 to 11 boxes: 23 per box 12 or more boxes: 22 per box with an additional 10% discount on the total paintball cost. Additionally, John's equipment maintenance costs 40 per month during the regular season and increases to 60 per month during the peak season. Travel costs to the paintball field vary each month (assume 4 weeks per month) as follows: Week 1: 10 Week 2: 15 Week 3: 12 Week 4: 8 Considering the changing frequency of playing, varying price per box, maintenance, and travel costs each month, calculate John's total monthly spending, including discounts, on paintball during the regular and peak seasons separately.
answer:Let's start by calculating John's monthly spending during the regular season. **Regular Season:** Paintball frequency: 3 times a month Boxes of paintballs per play: 3 boxes Total boxes per month: 3 plays * 3 boxes/play = 9 boxes Since John buys between 6 to 11 boxes, the cost per box is 23. Total paintball cost: 9 boxes * 23/box = 207 Equipment maintenance cost: 40 per month Travel costs per month: 10 + 15 + 12 + 8 = 45 Total monthly spending during the regular season: Paintball cost + Maintenance cost + Travel cost = 207 + 40 + 45 = 292 **Peak Season:** Paintball frequency: 3 times * 2 = 6 times a month Boxes of paintballs per play: 3 boxes Total boxes per month: 6 plays * 3 boxes/play = 18 boxes Since John buys 12 or more boxes, the cost per box is 22, with an additional 10% discount on the total paintball cost. Total paintball cost before discount: 18 boxes * 22/box = 396 Discount: 10% of 396 = 39.6 Total paintball cost after discount: 396 - 39.6 = 356.4 Equipment maintenance cost during peak season: 60 per month Travel costs per month remain the same: 45 Total monthly spending during the peak season: Paintball cost + Maintenance cost + Travel cost = 356.4 + 60 + 45 = 461.4 So, John's total monthly spending on paintball is boxed{292} during the regular season and 461.4 during the peak season.
question:ABCD is a cyclic quadrilateral. E and F are variable points on the sides AB and CD respectively, such that AE/EB = CF/FD. P is a point on the segment EF such that EP/PF = AB/CD. Show that area of triangle APD over the area of triangle BPC does not depend on the positions of E and F.
answer:Let ABCD be a cyclic quadrilateral with E and F points on sides AB and CD respectively, such that frac{AE}{EB} = frac{CF}{FD}. We need to show that the ratio of the areas of triangles triangle APD and triangle BPC does not depend on the positions of E and F. 1. **Step 1: Define variables and conditions** Let P be a point on the segment EF such that: [ frac{EP}{PF} = frac{AB}{CD} ] 2. **Step 2: Show fundamental triangles similarity** We first consider the case where AD and BC are not parallel. Assume that AD and BC intersect at point S. Since frac{AE}{EB} = frac{CF}{FD}, we know that triangle ASE sim triangle CSF (by the condition of similar triangles having the same angle and side ratio). 3. **Step 3: Angles and ratios** With triangle ASE sim triangle CSF, we have: [ angle DSE = angle CSF ] Additionally, from frac{EP}{PF} = frac{SA}{SC} (due to similar triangles and the given ratio), we understand that P is such that: [ frac{SE}{SF} = frac{SA}{SC} = frac{EP}{PF} ] 4. **Step 4: Angle bisectors in similar triangles** Since the triangles triangle ASE and triangle CSF are similar, and considering P's place on these ratios, line segment PS will bisect angle ESF. Therefore, we have: [ angle ASP = angle ASE - angle ESP ] [ angle CSP = angle CSF - angle PSF ] This implies that: [ angle ASP = angle CSP ] Hence, P lies on the common angle bisector. 5. **Step 5: Proving ratio independence** Since P lies on the angle bisector of angle ASB, the point is equidistant to lines AD and BC. Therefore: [ frac{text{area}(triangle APD)}{text{area}(triangle BPC)} = frac{AD}{BC} ] 6. **Step 6: Parallel case consideration** Suppose now that AD and BC are parallel. Since AB = CD (property of equal slopes in cyclic quadrilaterals), the point P will be the midpoint of EF. Given frac{AE}{EB} = frac{CF}{FD}, we reach the equality condition: [ left(frac{AB - BE}{BE}right) = left(frac{CD - DF}{DF}right) ] Consequently, BE = DF when AB = CD. Consider the perpendicular line L from P, which will equally cut through the positions: [ text{projection of } PF text{ on } L + text{projection of } DF text{ on } L = text{projection of } PE text{ on } L + text{projection of } BE text{ on } L ] 7. **Conclusion:** By demonstrating ratio independence for AD not being parallel and proving it naturally extends when AD parallel BC, we conclude: [ frac{text{area}(triangle APD)}{text{area}(triangle BPC)} = frac{AD}{BC} ] Therefore, the ratio of the area of triangle APD to triangle BPC does not depend on the positions of points E and F. (boxed{frac{text{area}(triangle APD)}{text{area}(triangle BPC)} = frac{AD}{BC}})