Ellipse practice · Medium

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 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 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.

Option D ($be\sqrt {{a^2} - {\alpha ^2}}$) is correct because it represents the area of the triangle formed by the foci and a point on the ellipse, derived using the properties of the ellipse.

Option A ($\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{e x}{a}$) is correct because it represents the locus of the midpoints of the focal chords of the ellipse.

Option B ($\alpha, \alpha+\frac{2 \pi}{3}, \alpha+\frac{4 \pi}{3}$) is correct because the area of the triangle is maximum when the eccentric angles are equally spaced around the ellipse, which occurs when they are $\alpha, \alpha+\frac{2 \pi}{3}, \alpha+\frac{4 \pi}{3}$.

Option B is correct because it directly applies the harmonic mean relationship for focal chords of an ellipse.

Option C ($\frac{4}{\sqrt{5}}$) is correct because after finding the eccentricity and semi-major axis, the latus rectum can be calculated using the formula $\frac{2b^2}{a}$.

Option A ($\left(\frac{-30}{23}, \frac{9}{23}\right)$) is correct because it represents the midpoint of the chord intercepted on the line by the ellipse.

The correct answer is D. The locus of the points of intersection of the tangents is an ellipse, and the length of its latus rectum is calculated to be 2 units.

The correct answer is $\mathrm{[1,3]and[1,3]}$ because the equation represents an ellipse, and the ranges for x and y can be determined from the standard form of the ellipse equation.

Option A ($b^4\left(c^2+a^2-b^2\right) x^2+c^2 a^4 y^2=a^4 b^4$) is correct because it represents the locus of P derived from the product of perpendiculars from the foci to the polar of P being equal to $c^2$.

Option C ($\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{a^2}}}$) is correct because it represents the locus of the poles of tangents to the auxiliary circle with respect to the given ellipse.

Option A ($a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=p^2$) is correct because it represents the condition for the line to be tangent to the ellipse, derived from equating the perpendicular distance from the center to the line with the condition for tangency.

Option D ($y=\frac{{3\sqrt{{5}}}}{{10}}$) is correct because it is derived by finding the equation of the ellipse and then finding the equation of the directrix.

The correct answer is D ($4,6$) because the value of b must be outside this interval for the ellipses to intersect at four distinct points.

Option A ($\pi / 4$) is correct because substituting the coordinates of a point on the ellipse into the distance formula yields $\theta = \pi/4$.

Option A (2) is correct because the discriminant of the quadratic equation formed by substituting the tangent equation into the ellipse equation is positive, indicating two real tangents.

Option C ($3 / 4$) is correct because the eccentricity of the ellipse can be calculated from the given equation by first converting it to standard form and then using the formula for eccentricity.

Option C ($\frac{{x^2}}{{a^2}} + \frac{{y^2}}{{b^2}} = \frac{{1}}{{2}}$) is correct because the chord PD touches this ellipse when CP and CD are conjugate semi-diameters.

Option B ($\frac{2}{\sqrt{7}}$) is correct because it satisfies the condition that the line segment LM subtends a right angle at the center of the auxiliary circle, which is derived by homogenizing the equation of the circle with the tangent line.

Option B ($a^2+6 a b \geq 7 b^2$) is correct because it represents the condition derived from the discriminant of the quadratic equation in $\alpha$ being non-negative.

Option B (2) is correct because it is derived from the harmonic mean relationship for focal chords.

Option D ($- \frac{{{b^2}}}{{{a^2}}}$) is correct because it is the relationship between the tangents of half angles of the parameters of the points on the ellipse that subtend a right angle at the vertex.

Option A is correct because by equating the tangent to the parabola and the normal to the ellipse, we can solve for both t and theta.

Option D ($x^2-y^2=a^2-b^2$) is correct because it represents the locus of the intersection point of tangents to the ellipse which make complementary angles with the x-axis.

Option B ($\frac{x^2}{4}+\frac{y^2}{3}=1$) is correct because it represents an ellipse with the given eccentricity and foci.