Hyperbola practice · Easy

Option C ($8 \sqrt{2}$) is correct because the differential equation leads to the equation of a rectangular hyperbola, and the length of the latus rectum for xy=c² is $2c\sqrt{2}$.

Option B ($\frac{2}{5}(2 x-y+4)^2-\frac{3}{5}(x+2 y-3)^2=1$) is correct because it correctly uses the given transverse and conjugate axes equations and their lengths to form the hyperbola equation.

Option B ((2,2)) is correct because by equating the parametric equations and solving for the intersection point, we find that x=2 and y=2.

Option D (${x^2} + {y^2} = 12$) is correct because the equation of the director circle of a hyperbola is given by ${x^2} + {y^2} = {a^2} - {b^2}$.

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

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 ($a^2(y^2-x^2)=4 x^2 y^2$) 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 (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 B ($t^2=2+\sqrt{5}$) is correct because it provides a real value for t that satisfies the hyperbola equation.

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 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.

Option B ($a^2\left(e^4 \sec ^2 \theta-1\right)$) is correct because it represents the product of the distances $AG$ and $A'G$, calculated using the coordinates of the vertices and the point where the normal intersects the transverse axis.