Ellipse practice · Easy

Option A ($\frac{\sqrt{5}-1}{2}$) is correct because it is the eccentricity calculated from the given condition that the distance between the foci equals the length of the latus rectum.

Option C ($x+2 y-4=0$) is correct because it represents the equation of the chord bisected at the given point.

Option B (Ellipse) is correct because the locus of the pole is an ellipse, derived by applying the tangency condition to the polar and the given curve.

Option A (8 / 3) is correct because the latus rectum is calculated using the formula 2b²/a after rewriting the equation in standard ellipse form.

Option D ($\frac{3}{5}$) is correct because the eccentricity of the ellipse is calculated using the formula $e = \sqrt{1 - \frac{a^2}{b^2}}$ after rewriting the equation in standard form.

Option C (${a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha = {p^2}$) is correct because it represents the condition for the given line to be a tangent to the ellipse.

Option D (None of these) is correct because the calculated tangent equation does not match any of the provided options.

The correct answer is C (${45^0}$) because it represents the eccentric angle of the point of contact of the line and the ellipse.

Option A (Ellipse) is correct because the given parametric equations can be manipulated to form the equation of an ellipse.

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

Option 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 ($4x - 3y = \pm 6\sqrt 5$) 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 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 A ($a b$) is correct because the maximum area of the triangle is achieved when P is at the end of the minor axis, resulting in an area of $ab$.