Ellipse

We have,
is an isosceles right angle triangle.

We are given the equation of the ellipse: The foci of the ellipse lie along the X-axis, and the distance between the foci is given as 8. Step 1: Use the formula for the distance between the foci The distance between the foci of the ellipse is given by , where . Here, (since ), and we are told the distance between the foci is 8. Thus: Now we can find using the equation : Squaring both sides: Solving for : Step 2: Apply the information about point We are also given that , where is the focus at , and is a point on the ellipse. We know that the distance from the point to the focus is . Using the parametric equations for an ellipse, the coordinates of a point on the ellipse are given by: For , the distance is given by: Substitute and : We are told that : Squaring both sides: Expanding the terms: Since , substitute this: Simplify: Rearrange: Solve the quadratic equation for using the quadratic formula: Thus, , so . Thus, the correct value of is . \bigskip

To solve this problem, we first need to understand the given equation of the ellipse . We rearrange it to its standard form: From this, we can identify the semi-major axis and semi-minor axis . For an ellipse, the equation for the latus rectum is given by , where is the distance from the center to the focus. Using the properties of the latus rectum and normal to the ellipse at a given point, we can derive the value of . Solving the equations and applying the conditions, we find the value of . Thus, the correct answer is .

We are given the equation of the ellipse , where . The length of the latus rectum is given as , and the distance from the center of the ellipse to its focus is . Step 1: Use the formula for the length of the latus rectum The length of the latus rectum of an ellipse is given by the formula: We are told that the length of the latus rectum is . Therefore, we have the equation: Solving for , we get: Step 2: Use the relationship between and The distance from the center of the ellipse to its focus is given by , and we are told that this distance is . Therefore, we have the equation: Squaring both sides: Substitute into this equation: Multiply through by 3 to eliminate the fraction: Rearranging this equation: Step 3: Solve the quadratic equation We can solve the quadratic equation using the quadratic formula: Thus, we have two possible solutions for : or Since and the value of must be positive, we take . Step 4: Find Substitute into the equation : So, . Step 5: Calculate We need to find . Substitute and : Thus, the correct answer is . \bigskip

We are given the hyperbola equation: and the point . Step 1: Equation of the tangent to the hyperbola For the hyperbola , the equation of the tangent at the point is: For the given hyperbola , we can write it in the standard form as: Thus, and . The equation of the tangent to the hyperbola at any point will be: Step 2: Slope of the tangent The slope of the tangent line at any point is given by . Using the values and , we have: Step 3: Using the point Now, we know that the tangent line passes through . So, the equation of the tangent line at is: Simplifying: Multiplying through by 4 to eliminate fractions: Thus, the relation between and is . Step 4: Solving for the values of and Now, substituting this relation into the equation of the hyperbola , we substitute into this equation. Expanding the terms: Simplifying: Rearranging: Dividing through by -2: Solving this quadratic equation using the quadratic formula: Substitute into : Step 5: Calculating the slope Finally, the slope of the tangent at is: After simplification: Thus, the correct value for the slope is . \bigskip

To find the equations of the directrices of the ellipse, we first write the ellipse equation in its standard form by completing the square. The directrices of an ellipse with a vertical major axis are given by the equation , where is the semi-major axis and is the eccentricity. Using the properties of the ellipse, we find that the directrices are . Thus, the correct answer is .