Ellipse practice · Easy

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.

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.

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.

Option B ($\frac{{x^2}}{{16}}+\frac{{y^2}}{{4}}=1$) is correct because it satisfies the condition that the line intersects the ellipse at points with eccentric angles 30° and 60° and subtends a right angle at the origin.

Option D ($2\sqrt{3}$) is correct because it represents the length of the minor axis calculated using the relationship between the semi-major axis, eccentricity, and semi-minor axis of the ellipse.

Option C ($x^2+y^2=13$) is correct because the locus of the point of intersection of perpendicular tangents to an ellipse is the director circle, given by $x^2 + y^2 = a^2 + b^2$.

Option A (4) is correct because the value of 'a' in the ellipse's equation is 4, derived from the foci and eccentricity.

Option A (2) is correct because it satisfies the condition that the extremities of the latus rectum lie on the given parabola.

Option B ($|t| \leq 1$) is correct because for the line to intersect the ellipse in real points, the condition $1-t^4 \geq 0$ must be satisfied, which implies $|t| \leq 1$.

Option D ($\frac{9}{2}$) is correct because the latus rectum of the ellipse is calculated using the formula $\frac{{2b^2}}{a}$ after rewriting the equation in standard form.

Option C ($\frac{\pi}{3}$) is correct because it is the value of $\theta$ that satisfies the given relationship between the eccentricities of the hyperbola and the ellipse.

Option B ($3y - 8x + 7 = 0$) is correct because substituting the point into the general normal equation and simplifying yields this result.

Option C ($\frac{{\sqrt 3 }}{2}$) is correct because it is the calculated eccentricity of the given ellipse after completing the square and putting it in standard form.

Option B ($\frac{{\pi}}{{6}}$) is correct because it minimizes the sum of the intercepts on the axes made by the tangent.

Option C ($\frac{1}{2}$) is correct because the eccentricity of the ellipse $E_2$ is calculated using the given information about the ellipses and the rectangle.

Option D ($\frac{{(x-2)^2}}{{9}}+\frac{{(y-1)^2}}{{25}}=1$) is correct because the locus is an ellipse with foci (2,-3) and (2,5), major axis 10, and center (2,1).

Option C ($\left(\frac{{3}}{{2}}, 3\right)$) is correct because the condition for the tangents to form an equilateral triangle leads to a range of values for the semi-major axis 'a' of the ellipse.