Ellipse practice · Easy

Option B ($\frac{\sqrt{3}}{2}$) is correct because it is the eccentricity calculated from the given condition that the latus rectum is half the minor axis.

Option A ($3{x^2} + 4{y^2} - 42x + 120 = 0$) is correct because it represents the equation of an ellipse with the given eccentricity and vertices.

Option D ($\frac{{{{\left( {x - 3} \right)}^2}}}{{25}} + \frac{{{{\left( {y - 2} \right)}^2}}}{{24}} = 1$) is correct because it satisfies the given foci and major axis length for an ellipse centered at (3, 2).

Option B ($\frac{{\sqrt 2 }}{3}$) is correct because by finding the equation of the normal to the ellipse and using the given intercepts, we can determine the value of $\beta$.

Whose eccentricity is $\sqrt{{\frac{{19}}{{3}}}}$ is the correct answer because after completing the square and rewriting the equation in standard hyperbola form, the eccentricity can be calculated using the formula $e = \sqrt{{1 + (b^2/a^2)}}$.

Option A ($\frac{{\sqrt 3 }}{2}$) is correct because the eccentricity is calculated by first rewriting the conic equation into standard ellipse form and then applying the eccentricity formula.

Option C ((1,1) ;(3,1)) is correct because, after completing the square and transforming the equation to standard form, the foci can be found using the eccentricity and center of the ellipse.

Option C ($5 / 13$) is correct because the eccentricity is calculated using the distance between the foci and the auxiliary circle property.

Option C ($c > \mathrm{max} \{a, b\}$) is correct because the radius of the circle must be greater than both the semi-major and semi-minor axes of the ellipse for the circle to contain the ellipse.

The correct answer is D ($e^4+e^2-1=0$) because it is derived from the condition that the normal at one end of the latus rectum passes through one end of the minor axis.

Option D (4) is correct because the number of points from which perpendicular tangents can be drawn is equal to the number of intersection points between the director circle of the inner ellipse and the outer ellipse.

Option A (${a^2} = 16,{b^2} = 12$) is correct because it correctly calculates the values of a² and b² based on the given foci and eccentricity.

Option B (abe) is correct because the maximum area of the triangle is obtained when the point P is at the end of the minor axis.

Option D ($\frac{{a^2 x^2}}{{\alpha^4}} + \frac{{b^2 y^2}}{{\beta^4}} = 1$) is correct because the locus of the poles is found by equating the polar equation to the tangent equation and using the condition for tangency.

Option B ((2,3)) is correct because the coordinates of P are found by equating the given equation of QR with the equation of the chord of contact.

Option D (${x^2} + {y^2} = 13$) is correct because the locus of the point of intersection of perpendicular tangents to an ellipse is its director circle, and its equation is ${x^2} + {y^2} = {a^2} + {b^2}$.

Option A ($3 x^2+4 y^2-42 x+120=0$) is correct because it is derived from the given eccentricity and vertices to form the equation of the ellipse.

Option C ($36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$) is correct because it represents the locus of the midpoints of the line segments, derived using parametric representation of the ellipse and trigonometric identities.

Option C is correct because the foci of the conic section are found by rewriting the equation in standard form and calculating the eccentricity.

Option B ($b^2$) is correct because the product of perpendiculars from foci to any tangent of an ellipse equals the square of the semi-minor axis.

Option B (27) is correct because it represents the area of the quadrilateral formed by the tangents at the endpoints of the latus recta of the ellipse.

Option A (9) is correct because by parameterizing the ellipse and maximizing the given expression, the largest possible value is found to be 9.

Option A ($2 a^2$) is correct because the sum of the squares of the perpendiculars simplifies to $2a^2$ after algebraic manipulation and using the ellipse properties.

Option C ($2ab\sin \frac{{\alpha - \beta }}{2}\sin \frac{{\beta - \gamma }}{2}\sin \frac{{\gamma - \alpha }}{2}$) is correct because it represents the area of the triangle formed by the given points on the ellipse using eccentric angles.

Option D ($r \in (2,5)$) is correct because for the equation to represent an ellipse, both denominators must be positive.

The correct answer is A ($10\sqrt 2$) because it represents the length of the major axis of the ellipse, calculated using the given distances between the foci and directrices.

Option C ($\left( {a\frac{{{e^t} + {e^{ - t}}}}{2},b\frac{{{e^t} - {e^{ - t}}}}{2}} \right)$) is correct because these parametric coordinates satisfy the equation of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$.

The correct answer is 6 because it satisfies the condition for conjugate points with respect to the given ellipse.

The correct answer is A ((-1,2) and (-1,-6)) because it correctly identifies the foci of the ellipse after translating the standard ellipse equation.

Option D (16) is correct because after calculating the eccentricities of the ellipse and hyperbola, substituting them into the given ellipse equation yields k=16.