Let for two distinct values of p the lines y=x+p touch the ellipse E:4x2+3y2=1 at the points A and B. Let the line y=x intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to:
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Step-by-Step Solution
Step 1: Find the tangent condition for the ellipse
The general condition for a line y=mx+c to be tangent to the ellipse a2x2+b2y2=1 is given by c2=a2m2+b2. For the given ellipse 4x2+3y2=1, we have a2=4 and b2=3. The tangent lines are y=x+p, so m=1 and c=p.
Step 2: Determine the values of p and tangent points A and B
Substituting the values into the tangent condition, we get p2=4(1)2+3, which simplifies to p2=7. Thus, the two distinct values for p are 7 and −7. The points of tangency (x0,y0) for the line y=mx+c are given by x0=−ca2m and y0=cb2.
Step 3: Calculate coordinates of A and B
For p=7, the point of tangency A is (−74(1),73)=(−74,73). For p=−7, the point of tangency B is (−−74(1),−73)=(74,−73).
Step 4: Find the intersection points C and D with line y=x
To find the intersection points C and D of the line y=x with the ellipse, substitute y=x into the ellipse equation: 4x2+3x2=1. This simplifies to 3x2+4x2=12, so 7x2=12. Solving for x, we get x=±712=±723. Since y=x, the coordinates are C=(723,723) and D=(−723,−723).
Step 5: Calculate the area of quadrilateral ABCD
The quadrilateral ABCD is symmetric with respect to the origin. The area of a quadrilateral with vertices (x1,y1),(x2,y2),(x3,y3),(x4,y4) can be calculated using the formula: 21∣(x1y2−y1x2)+(x2y3−y2x3)+(x3y4−y3x4)+(x4y1−y4x1)∣. Alternatively, we can use the fact that the quadrilateral is symmetric and its diagonals intersect at the origin. The area of a quadrilateral whose diagonals are perpendicular and bisect each other is 21d1d2. Here, the diagonals are AB and CD. The line y=x and y=−x are perpendicular. The line AB has slope −1/1=−1 (from y=x+p and y=−x+p). The line CD has slope 1. So the diagonals are perpendicular. The origin is the midpoint of both AB and CD. Thus, the area is 21×length(AB)×length(CD).
Step 6: Calculate lengths of diagonals and final area
The length of diagonal AB is (78)2+(−76)2=764+736=7100=710. The length of diagonal CD is (−743)2+(−743)2=748+748=796=746. The area of the quadrilateral ABCD is 21×length(AB)×length(CD)=21×710×746=7206.