The general condition for a line y = mx + c to be tangent to the ellipse (x²)/(a²) + (y²)/(b²) = 1 is given by c² = a²m² + b². For the given ellipse (x²)/(4) + (y²)/(3) = 1, we have a² = 4 and b² = 3. The tangent lines are y = x + p, so m = 1 and c = p.

The general condition for a line y = mx + c to be tangent to the ellipse (x²)/(a²) + (y²)/(b²) = 1 is given by c² = a²m² + b². For the given ellipse (x²)/(4) + (y²)/(3) = 1, we have a² = 4 and b² = 3. The tangent lines are y = x + p, so m = 1 and c = p.

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 (x_0, y_0) for the line y = mx + c are given by x_0 = -(a²m)/(c) and y_0 = (b²)/(c).

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 (x_0, y_0) for the line y = mx + c are given by x_0 = -(a²m)/(c) and y_0 = (b²)/(c).

For p = √7, the point of tangency A is (-(4(1))/(√7), (3)/(√7)) = (-(4)/(√7), (3)/(√7)). For p = -√7, the point of tangency B is (-(4(1))/(-√7), (3)/(-√7)) = ((4)/(√7), -(3)/(√7)).

For p = √7, the point of tangency A is (-(4(1))/(√7), (3)/(√7)) = (-(4)/(√7), (3)/(√7)). For p = -√7, the point of tangency B is (-(4(1))/(-√7), (3)/(-√7)) = ((4)/(√7), -(3)/(√7)).

To find the intersection points C and D of the line y = x with the ellipse, substitute y = x into the ellipse equation: (x²)/(4) + (x²)/(3) = 1. This simplifies to 3x² + 4x² = 12, so 7x² = 12. Solving for x, we get x = ±(√12)/(√7) = ±(2√3)/(√7). Since y = x, the coordinates are C = ((2√3)/(√7), (2√3)/(√7)) and D = (-(2√3)/(√7), -(2√3)/(√7)).

To find the intersection points C and D of the line y = x with the ellipse, substitute y = x into the ellipse equation: (x²)/(4) + (x²)/(3) = 1. This simplifies to 3x² + 4x² = 12, so 7x² = 12. Solving for x, we get x = ±(√12)/(√7) = ±(2√3)/(√7). Since y = x, the coordinates are C = ((2√3)/(√7), (2√3)/(√7)) and D = (-(2√3)/(√7), -(2√3)/(√7)).

The length of diagonal AB is ((8)/(√7))² + (-(6)/(√7))² = √((64)/(7) + (36)/(7)) = √((100)/(7)) = (10)/(√7). The length of diagonal CD is (-(4√3)/(√7))² + (-(4√3)/(√7))² = √((48)/(7) + (48)/(7)) = √((96)/(7)) = (4√6)/(√7). The area of the quadrilateral ABCD is (1)/(2) × length(AB) × length(CD) = (1)/(2) × (10)/(√7) × (4√6)/(√7) = (20√6)/(7).

The length of diagonal AB is ((8)/(√7))² + (-(6)/(√7))² = √((64)/(7) + (36)/(7)) = √((100)/(7)) = (10)/(√7). The length of diagonal CD is (-(4√3)/(√7))² + (-(4√3)/(√7))² = √((48)/(7) + (48)/(7)) = √((96)/(7)) = (4√6)/(√7). The area of the quadrilateral ABCD is (1)/(2) × length(AB) × length(CD) = (1)/(2) × (10)/(√7) × (4√6)/(√7) = (20√6)/(7).

Let for two distinct values of $p$ the lines $y = x + p$ touch the ellipse $E:\;\frac{x^2}{4}+\frac{y^2}{3}=1$ 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:

Answer: $\frac{20\sqrt{6}}{7}$

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 $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $c^2 = a^2m^2 + b^2$ . For the given ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$ , we have $a^2 = 4$ and $b^2 = 3$ . 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 $p^2 = 4(1)^2 + 3$ , which simplifies to $p^2 = 7$ . Thus, the two distinct values for $p$ are $\sqrt{7}$ and $-\sqrt{7}$ . The points of tangency $(x_0, y_0)$ for the line $y = mx + c$ are given by $x_0 = -\frac{a^2m}{c}$ and $y_0 = \frac{b^2}{c}$ .

Step 3: Calculate coordinates of A and B

For $p = \sqrt{7}$ , the point of tangency $A$ is $\left(-\frac{4(1)}{\sqrt{7}}, \frac{3}{\sqrt{7}}\right) = \left(-\frac{4}{\sqrt{7}}, \frac{3}{\sqrt{7}}\right)$ . For $p = -\sqrt{7}$ , the point of tangency $B$ is $\left(-\frac{4(1)}{-\sqrt{7}}, \frac{3}{-\sqrt{7}}\right) = \left(\frac{4}{\sqrt{7}}, -\frac{3}{\sqrt{7}}\right)$ .

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: $\frac{x^2}{4} + \frac{x^2}{3} = 1$ . This simplifies to $3x^2 + 4x^2 = 12$ , so $7x^2 = 12$ . Solving for $x$ , we get $x = \pm\frac{\sqrt{12}}{\sqrt{7}} = \pm\frac{2\sqrt{3}}{\sqrt{7}}$ . Since $y = x$ , the coordinates are $C = \left(\frac{2\sqrt{3}}{\sqrt{7}}, \frac{2\sqrt{3}}{\sqrt{7}}\right)$ and $D = \left(-\frac{2\sqrt{3}}{\sqrt{7}}, -\frac{2\sqrt{3}}{\sqrt{7}}\right)$ .

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 $(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)$ can be calculated using the formula: $\frac{1}{2} |(x_1y_2 - y_1x_2) + (x_2y_3 - y_2x_3) + (x_3y_4 - y_3x_4) + (x_4y_1 - y_4x_1)|$ . 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 $\frac{1}{2}d_1d_2$ . 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 $\frac{1}{2} \times \text{length}(AB) \times \text{length}(CD)$ .

Step 6: Calculate lengths of diagonals and final area

The length of diagonal $AB$ is $\sqrt{\left(\frac{8}{\sqrt{7}}\right)^2 + \left(-\frac{6}{\sqrt{7}}\right)^2} = \sqrt{\frac{64}{7} + \frac{36}{7}} = \sqrt{\frac{100}{7}} = \frac{10}{\sqrt{7}}$ . The length of diagonal $CD$ is $\sqrt{\left(-\frac{4\sqrt{3}}{\sqrt{7}}\right)^2 + \left(-\frac{4\sqrt{3}}{\sqrt{7}}\right)^2} = \sqrt{\frac{48}{7} + \frac{48}{7}} = \sqrt{\frac{96}{7}} = \frac{4\sqrt{6}}{\sqrt{7}}$ . The area of the quadrilateral $ABCD$ is $\frac{1}{2} \times \text{length}(AB) \times \text{length}(CD) = \frac{1}{2} \times \frac{10}{\sqrt{7}} \times \frac{4\sqrt{6}}{\sqrt{7}} = \frac{20\sqrt{6}}{7}$ .

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Let for two distinct values of ppp the lines y=x+py = x + py=x+p touch the ellipse E: x24+y23=1E:\;\frac{x^2}{4}+\frac{y^2}{3}=1E:4x2​+3y2​=1 at the points AAA and BBB. Let the line y=xy = xy=x intersect EEE at the points CCC and DDD. Then the area of the quadrilateral ABCDABCDABCD is equal to: