Ellipse practice · Mixed difficulty

Option B ($c^2(a^4 x^2+b^4 y^2)=a^4 b^4$) is correct because the polar of the ellipse touches the circle, so the condition for tangency can be used to find the locus of the pole.

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 A (($1,2$) and ($- 3,2$)) is correct because these are the coordinates of the foci after completing the square and finding the center and eccentricity.

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 C ((0,0)) is correct because the circle on $TT'$ as diameter passes through the origin.

Option B ($\sqrt{2}-1$) is correct because the eccentricity is derived from the relationship between the distance between foci and the latus rectum length.

Option A ($a^2 l^2+b^2 m^2=n^2$) is correct because this is the condition for the line $lx + my - n = 0$ to be tangent to the ellipse $\frac{{x^2}}{{a^2}} + \frac{{y^2}}{{b^2}} = 1$.

Option B (ellipse) is correct because the locus of a point with a constant ratio of distances from two fixed points (less than 1) defines an ellipse.

Option C (10) is correct because PF1 + PF2 equals the length of the major axis, which is 2a, and a is 5 in this ellipse equation.

Option C ($3{x^2} + 5{y^2} = 32$) is correct because it represents the equation of an ellipse that passes through the point (-3, 1) and has the given eccentricity.

Option B ($\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{\left(a^2-b^2\right)^2}{n^2}$) is correct because it represents the condition for the line to be normal to the ellipse.

The correct answer is B ($\frac{2}{\sqrt{5}}$) because it is derived from the relationship between the minor axis, distance between foci, and eccentricity of an ellipse.

Option D (9/2) is correct because the length of the latus rectum of the ellipse is calculated using the formula 2b²/a, where a and b are the semi-major and semi-minor axes, respectively.

Option A ($6 \sqrt{5}$ and $4 \sqrt{5}$) is incorrect. The correct lengths of the major and minor axes are 6 and 4, respectively, obtained by simplifying the ellipse equation.

Option A (192) is correct because the eccentricities are reciprocals, leading to the equation 1 + b²/64 = 4, which solves to b² = 192.

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.

Option D ($3 / 5$) is correct because it is the calculated eccentricity of the given ellipse after converting it to standard form.

Option C ($2 \sqrt{8}, 3 \sqrt{2}$) is correct because these are the dimensions of the rectangle of maximum area inscribed in the given ellipse.

Option D ($5{x^2} + 9{y^2} = 180$) is correct because it represents the ellipse with the given distance between foci and directrices.

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 A ($x+y=4$) is correct because it represents the line joining the foci of the two given ellipses.

Option A ($\left(-1, \frac{1}{2}\right)$) is correct because it is found by comparing the given polar equation with the general polar equation T=0.

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 (4) is correct because the area of the triangle can be calculated using the coordinates of the foci and a point on the ellipse, given the relationship between the distances from the point to the foci.

Option C ($x^2+y^2=a^2+b^2+\lambda$) is correct because it represents the locus of the intersection point of perpendicular tangents to the two given ellipses.

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.