Ellipse practice · Mixed difficulty

Option B (-$\frac{2}{3}$) is correct because it is the solution to the equation derived from the intersection of the normal and the ellipse.

Option A ($\frac{a^2}{\alpha^2}+\frac{b^2}{\beta^2}=2$) is correct because it represents the relationship between the semi-major and semi-minor axes of the two ellipses when their intersection points are at the extremities of conjugate diameters.

Option B (2 / 3) is correct because, after rewriting the equation in standard form, the eccentricity is calculated using the formula $e = \sqrt{{\frac{{b^2 - a^2}}{{b^2}}}}$ since b > a.

Option B ($3(x-3 y+3)^2+2(3 x+y-1)^2=180$) is correct because it represents the equation of the ellipse with given axes lengths and equations.

Option B ($\frac{16}{5}$) is correct because it represents the minimum distance from the center of the ellipse to the chord of contact of mutually perpendicular tangents, which is related to the radius of the director circle.

Option D ($\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}$) is correct because it represents the locus of point P such that the chord of contact subtends a right angle at the center of the ellipse.

The correct answer is $\sqrt {\frac{6}{7}}$ because it is the eccentricity calculated from the given equation of the ellipse.

Option A ($15\pi$) is correct because the area of the elliptical path is calculated using the formula $A = \pi ab$, where a and b are the semi-major and semi-minor axes, respectively.

Option D ($\left(\frac{4 a m^2}{l^2}, \frac{4 a m}{l}\right)$) is correct because it represents the coordinates of point D, which is the intersection of the third normal from point C with the parabola.

Option B ($\ - 5\\,\pi /6$) is correct because it represents the eccentric angle of point B, which is diametrically opposite to point A.

Option D is correct because the condition for four normals to be drawn from a point (h,0) to the ellipse is |h| < ae².

Option C (7) is correct because the value of b² is calculated by equating the foci of the ellipse and hyperbola.

The correct answer is 1 because the given condition implies a relationship between the eccentricity and the semi-major/minor axes, leading to ${e^4} + {e^2} = 1$.

Option A ($- 15 \le F \le 15$) is correct because the point $(\alpha, \beta)$ lies on the director circle of the ellipse, and by expressing $\alpha$ and $\beta$ in terms of trigonometric functions, we can find the range of $F$.

The correct answer is C ($3y = \pm 25$) because it represents the equation of the directrix of the given ellipse.

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

Option A (2) is correct because the radius of the largest circle that can be inscribed in the ellipse with its center at one focus is equal to a - ae, where a is the semi-major axis and e is the eccentricity.

Option C ($5 x+4 y=13$) is correct because it is obtained by using the equation $T=S_1$ for the chord bisected at a given point.

Option B ($3 \sqrt{2}$) is correct because it is the length of the latus rectum calculated from the given distance between foci and directrices.

Option D ($y$-axis) is correct because the normal to the ellipse at the point (0,3) is along the y-axis.

Option C (-4) is correct because it is found using the condition $c = -a m^2$ for tangency.

The correct answer is B (8) because it represents the sum of the distances from any point P on the ellipse to the two foci A and B, which is equal to 2a, where a is the semi-major axis.

Option C ($\pm \frac{\pi }{3}$) is correct because it represents the value of $\phi$ for which the tangent to the ellipse is also a tangent to the circle, found by equating the distance from the center of the circle to the tangent line with the radius of the circle.

Option D ((-1,1)) is correct because it represents the midpoint of the chord formed by the intersection of the given line and conic.

Option D ($a \in(-1,5)$) is correct because the condition for the major axis to be the Y-axis leads to the inequality $a^2-5 < 4a$, which gives the specified interval.

Option D ($5 / 13$) is correct because the eccentricity is calculated using the distance between foci and the property that the foot of the perpendicular from the focus to a tangent lies on the auxiliary circle.

Option B ($3 \sqrt{5}$) is correct because it is the area of the triangle formed by the point on the parabola and the two points on the ellipse, calculated using coordinate geometry.

Option A (${b^2}$) is correct because the product of CQ and CR is equal to the square of the semi-minor axis of the ellipse.

Option C (On the major axis) is correct because substituting the point (1,3) into the ellipse equation results in a negative value, indicating the point lies inside the ellipse, specifically on the major axis.

Option A, $\frac{e^2}{\sqrt{1-e^2}}$, is the correct expression for $\frac{2 \cos \omega}{\sin 2 \theta}$ derived from the angle between the normals to the ellipse.