Ellipse practice · Mixed difficulty

Option B ($\left| t \right| \le 1$) is correct because the condition for real points requires the y-coordinate to be real.

Option B ($\frac{10}{3}$) is correct because the length of the transverse axis is calculated using the eccentricity and the fact that the focal chord passes through a focus.

Option C ($\frac{1}{{\sqrt 2 }}$) is correct because after finding the locus of the midpoint and identifying it as an ellipse, its eccentricity is calculated as $\frac{1}{\sqrt{2}}$.

Option A (1) is correct because it represents the constant value of the expression derived from the properties of tangents to an ellipse.

Option D ($\sqrt{10}$) is correct because it represents the distance between the two points on the ellipse where the tangents are parallel to the given chord.

The correct answer is B. By finding the equation of the normal at the end of the latus rectum and using the given condition, the relationship ${e^2}(1 + {e^2}) = 1$ is derived.

Option C ($5 x^2+9 y^2=125$) is correct because it represents an ellipse passing through (5,0) with the given eccentricity.

Option D ($x + 1 = 0, y - 1 = 0$) is correct because completing the square gives the standard ellipse form, and the axes are lines through the center parallel to the coordinate axes.

Option B ($4x - 2y = 1$) is correct because it represents the equation of the normal to the ellipse at the given point, calculated using the properties of ellipses and their directrices.

Option B ($3 \sqrt{6}+6$) is correct because it represents the maximum area of the triangle inscribed in the given ellipse, calculated using parametric coordinates and maximizing the area function.

Option B ($\frac{{2a^2b^2}}{{a^2+b^2}}$) is correct because it is derived by finding the condition for a common tangent to the ellipse and the circle.

Option A ($y^{2}=-104 x$) is correct because the normal to the ellipse at the given point touches the parabola $y^2 = 4ax$ when the condition $a - mc = 0$ is satisfied.

Option A (16) is correct because the product of the perpendiculars from the foci to any tangent of the ellipse is equal to the square of the semi-minor axis.

Option B ($2 x-y=1$) is correct because it represents the equation of the normal to the ellipse at (2,3) calculated using the given conditions.

Option C ($\frac{{\sqrt{{7}}}}{{4}}$) is correct because the condition that the parallelogram's fourth vertex lies on the circumcircle implies a rectangle, leading to the director circle condition and allowing calculation of the eccentricity.

Option C ($\frac{{13}}{5}$) is correct because it represents the ratio a/b when the eccentricities of the two given ellipses are equal.

Option C ($\frac{{\sqrt{{3}}}}{{2}}$) is correct because it satisfies the ellipse equation after substituting the given points and solving for the relationship between a and b.

Option A ($\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{c^2}$) is correct because it represents the locus of the point from which the chord of contact of tangents to the ellipse touches the circle.

Option A ($\frac{(x-5)^2}{25}+\frac{y^2}{16}=1$) is correct because the given condition leads to the sum of distances from A to B and C being constant, which defines an ellipse with the given equation.

Option A ($4x - 3y = \pm 6\sqrt 5$) is correct because it represents the equations of tangents to the ellipse that are perpendicular to the given line.

Option B (12) is correct because the sum of the focal distances from any point on an ellipse is equal to the length of its major axis (2a).

The correct answer is B (centre) because the condition $\tan \theta_1 \tan \theta_2=-\frac{a^2}{b^2}$ implies that the lines joining the points to the center are perpendicular.

Option A ($\frac{{1}}{{2 x^2}}+\frac{{1}}{{4 y^2}}=1$) is correct because it represents the locus of the midpoints of the intercepts formed by the tangents to the ellipse.

Option C, $\sqrt{2(\sqrt{2}-1)}$, is the correct eccentricity value derived by using the equation of the normal to the ellipse and the given point.

Option C ($(–\infty, 2)$) is correct because for the equation to represent an ellipse, both denominators must be positive, leading to $a < 5$ and $a < 2$.

Option D ($\frac{\pi}{12}$) is correct because the angle between the common tangent and the given line is calculated using the tangent formula and simplifies to $\frac{\pi}{12}$.

Option A ($\frac{2 \sqrt{6}}{7}$) is correct because it is the eccentricity calculated from the given ellipse parameters after finding a and b.

Option C ($\frac{\pi }{4}$) is correct because the eccentric angle is found by equating the given line to the tangent equation of the ellipse and solving for the angle.

Option A ($\frac{5}{7}$) is correct because by finding the values of a and b from the given information, the eccentricity can be calculated using the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$.

Option D (4 sq. unit) is correct because the area of the triangle is calculated using the properties of the ellipse and the given ratio of distances to the foci.