Ellipse practice · Mixed difficulty

Option A ($\frac{{1}}{{2}}$) is correct because the eccentricity is calculated using the distances from a point on the ellipse to its foci, resulting in e = 1/2.

Option D (32) is correct because the length of the major axis (2a) is calculated using the distance between foci (2ae) and the eccentricity (e).

Option C (Ellipse) is correct because the ratio of the distance from a point to a fixed point (focus) and the distance to a fixed line (directrix) is constant and less than 1, which defines an ellipse.

Option D (None of these) is correct because the locus of the foot of the perpendicular from the center to any tangent of the ellipse is neither a circle, an ellipse, nor a hyperbola.

Option C ($(5, \infty)$) is correct because the ellipse equation requires both denominators to be positive, leading to the intersection of the solution sets $r < -2$ or $r > 3$ and $r < 1$ or $r > 5$.

Option B ($3{x^2} + 4{y^2} = 12$) is correct because given the eccentricity and directrix, we can determine the values of a and b, and thus the equation of the ellipse.

Option B ($\frac{{72}}{7}$) is correct because the length of the latus rectum of the ellipse is calculated as $\frac{{2{a^2}}}{b} = \frac{{72}}{7}$.

Option B (1) is correct because the tangent at any point (a cos θ, b sin θ) to the ellipse cuts off intercepts h and k such that $\frac{{a^2}}{{h^2}} + \frac{{b^2}}{{k^2}} = 1$.

Option B ($\frac{x^2}{81}+ \frac{y^2}{45}=\frac{1}{4}$) is correct because it satisfies the given conditions for the ellipse.

Option B ($4x + 5y = 13$) is correct because it uses the T = S' formula for the equation of the chord bisected at a given point.

Option C is correct because the equation of the ellipse is determined by finding the center, semi-major axis, and semi-minor axis from the given vertices and eccentricity.

Option C ($100 x^2+36 y^2=3600$) is correct because it is the equation of the ellipse derived from the given vertices and eccentricity, with the major axis along the y-axis.

Option B ($\frac{c-a}{c+a}$) is correct because it represents the relationship between the tangents of half angles and the distance c.

Option C ($\frac{{16}}{3}$) is correct. By substituting the given point and eccentricity into the ellipse equation, we find the values of $a^2$ and $b^2$. Then, using the focus and radius of the circle, we find the intersection points and calculate the square of the distance between them.

The correct answer is $\cos \theta$ because it represents the eccentricity of the ellipse, derived from the given angle and the properties of the ellipse.

Option A ($\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = \frac{1}{2}$) is correct because it represents the locus of the midpoint of PQ, derived using parametric coordinates and trigonometric identities.

Option D ($4/\sqrt 3$) is correct because it is calculated using the formula for the latus rectum of an ellipse.

Option D ($a / \sqrt{{2}}$) is correct because it represents the length of the subtangent to the ellipse at the given parameter value.

Option C ($\frac{{2\pi}}{{3}}$) is correct because it satisfies the condition where the tangent to the ellipse touches the given circle.

Option B ($\frac{1}{\sqrt{1+\sin ^2 \theta}}$) is correct because it represents the eccentricity of the ellipse when the chord QR subtends a right angle at the center C.

Option B (4) is correct because it involves finding the maximum area of a parallelogram inscribed in an ellipse, using the given eccentricity to determine the value of 'a'.

The correct answer is B ((-2,1)) because it represents the center of the given ellipse after completing the square and rewriting the equation in standard form.

Option D (20) is correct because it satisfies the condition for equal eccentricities of the two given ellipses.

Option C ($\left(1 \pm \frac{1}{\sqrt{6}}, 2\right)$) is correct because it represents the foci of the given ellipse after rewriting it in standard form and calculating the eccentricity.

Option A (1) is correct because it follows directly from the tangent equation to the ellipse and the intercepts formed on the coordinate axes.

Option C ($\frac{{3}}{{5}}$) is correct because the eccentricity is calculated using the formula $e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}}$ since $b > a$.

Option B ($\sqrt{a^2+b^2}$) is correct because setting the x and y intercepts of the tangent equal to $l$ and solving yields $l = \sqrt{a^2 + b^2}$.

Option C (4) is correct because the tangent to the ellipse being normal to the circle implies it passes through the circle's center, leading to a constraint on a and b, and maximizing ab gives 4.

Option C (2) is correct because the point (3,5) lies outside the ellipse, allowing two real tangents to be drawn.

Option D ($\frac{(x-2)^2}{9}+\frac{(y-1)^2}{8}=9$) is incorrect, the correct equation is $\frac{(x-2)^2}{72}+\frac{(y-1)^2}{81}=1$. There appears to be a mistake in the provided answer options.