Hyperbola practice · Mixed difficulty

Option B ($(6,3)$ and $(-2,3)$) is correct because these are the vertices found after completing the square and rewriting the hyperbola equation in standard form.

Option A ($x^2-16 x y-11 y^2-12 x+6 y+21=0$) is correct because it represents the equation of the hyperbola derived using the focus-directrix definition and the given eccentricity.

Option C ($30^{\circ}$) is correct because it is the angle that satisfies the condition where the tangent to the hyperbola is parallel to the given line.

Option C ($a^2(y^2-x^2)=4 x^2 y^2$) is correct because it represents the locus of the intersection point of tangents at the ends of a normal chord of the hyperbola.

Option B ($45^{\circ}$) is correct because substituting the point's coordinates into the distance formula and simplifying leads to $\tan^2 \theta = 1$, which gives $\theta = 45^{\circ}$ in the first quadrant.

Option A is correct because the equation of the hyperbola is found using the given distances of the vertex from the foci.

Option A (A hyperbola with length of its transverse axis $8 \sqrt{2}$) is correct because the locus is a hyperbola, and the transverse axis length is correctly calculated.

Option A ($\frac{x^2}{a^2 \sec ^2 \theta}-\frac{y^2}{b^2}=1$) is correct because the chord PQ touches this hyperbola.

Option B ($\sqrt{2}$) is correct because when the transverse and conjugate axes are equal (a=b), the eccentricity of the hyperbola is $\sqrt{2}$.

Option B ($\sqrt 2$) is correct because the eccentricity of a rectangular hyperbola, where a = b, is always $\sqrt 2$.

Option D ($\sqrt{3}$) is correct because it is the eccentricity calculated from the standard form of the given hyperbola equation.

Option C ($\frac{4 \sqrt{5}}{3}$) is correct because after finding the relationship between the eccentricities and using the given condition, we can determine the value of $\cos^2 \theta$ and then calculate the latus rectum of the ellipse.

Option A ($(4 \sqrt{2}, 3)$) is correct because the area of the triangle PQS is 27/10 sq. units for this point.

Option B ($\frac{2}{3}$) is correct because the product of the lengths of perpendiculars from a point on the hyperbola to its asymptotes is $\frac{a^2b^2}{a^2+b^2}$.

Option A (2) is correct. By using the given information about the directrix and latus rectum, the eccentricity can be found by solving the resulting equation.

Option B ($\left(x^2+y^2\right)^2=a^2 x^2-b^2 y^2$) is correct because the locus of the foot of the perpendicular from the center of the hyperbola on a variable tangent is given by this equation.

Option A (73) is correct because by substituting the point (c, d) into the tangent equation and using the given condition $\tan \alpha \tan \beta = 1$, we can solve for $c^2 - d^2$.

Option A (-8) is correct because the arithmetic mean of the intersection points of the circle and rectangular hyperbola is the midpoint of the centers of the hyperbola and circle, allowing us to solve for g+f.

Option B ($t^2=2+\sqrt{5}$) is correct because it provides a real value for t that satisfies the hyperbola equation.

Option D (8) is correct because the semi-transverse axis of a rectangular hyperbola $xy=c^2$ is given by $a = \sqrt{2}c$, and in this case, $c^2 = 32$.

Option A ($x - y + 1 = 0$) is correct because substituting the line equation into the hyperbola equation and using the tangency condition yields this result.

Option C ($5 x^2-4 y^2=80$) is correct because it represents a hyperbola with the given center, vertex, and focus.

Option A ($\sqrt {\frac{7}{3}}$) is correct because the eccentricity of the hyperbola is calculated correctly after rewriting the equation in standard form.

Option A ($x-y+1=0$) is correct because it represents a tangent line to the hyperbola with a slope of 1, derived using the tangent equation $c^2=a^2 m^2-b^2$.

Option C (A hyperbola) is correct because the difference of distances from two fixed points being constant defines a hyperbola.

Option A ($\frac{x}{{{x_1} + {x_2}}} + \frac{y}{{{y_1} + {y_2}}} = 1$) is correct because it represents the equation of the chord joining two points on the rectangular hyperbola, derived using the midpoint form of the chord equation.

Option B ($y^2-x^2=5$) is correct because it represents a hyperbola with foci at $(0, \pm \sqrt{10})$ and passing through the point $(2,3)$.

Option A ($\sec \lambda$) is correct because it represents the eccentricity of the hyperbola, derived from the relationship between the area of the triangle formed by the asymptotes and a tangent, and the hyperbola's parameters.

Option D (4) is correct because by comparing the given tangent equation with the standard form of a tangent to a hyperbola, we can derive the relationship between $\lambda$, $\mu$, a, and b, and find the value of the expression.

Option B ($\frac{{{{x^2}}}}{9} - \frac{{{{(y - 4)}^2}}}{7} = \frac{1}{4}$) is correct because it's the equation of the hyperbola derived from the given foci and eccentricity.