Hyperbola practice · Mixed difficulty

Option A ($y=3 x \pm \sqrt{6}$) is correct because it represents the equations of the tangents to the hyperbola that are perpendicular to the given line, calculated using the slope and the hyperbola's parameters.

Option A ($3{x^2} + 4{y^2} + 8xy - 8x - 4y - 3 = 0$) is correct because it represents a pair of straight lines and satisfies the condition for the asymptotes of the given hyperbola.

Option C (1) is correct because calculating the eccentricities $e_1$ and $e_2$ and then summing their reciprocals squared results in 1.

Option C ($9 x^2-25 y^2=225$) is the correct equation of the hyperbola because it satisfies the given asymptotes and vertices.

Option B (${a^2}$) is correct because the product of OT and ON simplifies to the square of the semi-major axis of the hyperbola.

Option D ($2{x^2} + 5xy + 2{y^2} + 4x + 5y + 2 = 0$) is correct because it represents the combined equation of the asymptotes of the given hyperbola, obtained by setting the determinant condition for a pair of straight lines to zero.

Option B ($3 x^2-y^2=27$) is correct because it represents the equation of a hyperbola with foci at $(\pm 6, 0)$ and eccentricity 2, derived from the relationship between a, b, and e.

Option D (A hyperbola) is correct because eliminating the parameter 't' from the given parametric equations results in the equation of a hyperbola.

Option C ($\frac{5}{4}$) is correct because the eccentricity of the conjugate hyperbola is calculated using the relationship between the eccentricities of a hyperbola and its conjugate.

Option D ($-\frac{13}{2}$) is correct because it is the ordinate of the intersection point of the normals at P and Q, derived using the properties of hyperbolas and trigonometric identities.

Option B (6) is correct because the difference of the distances from any point on the hyperbola to the foci is constant and equal to 2a.

Option B ($\pm 6$) is correct because it correctly uses the condition for tangency to find the value of $\lambda$.

Option C ($6\sqrt 3$) is correct because it represents the distance QR calculated using the properties of the hyperbola, its tangent, and normal.

Option B ($(3,\infty )$) is correct because the length of the latus rectum is greater than 3 when the eccentricity of the hyperbola is greater than 2.

Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the mean deviation is correctly calculated using the provided formula, and the reason accurately explains the method.

Option A (${y_1}{y_2}{y_3}{y_4} = 1$) is correct because the product of the y-coordinates of the intersection points of the circle and the rectangular hyperbola xy=1 is 1.

Option A ($3 x^2-y^2=3$) is correct because it correctly uses the foci and eccentricity to derive the hyperbola equation.

Option B ($\frac{3}{2}$) is correct because after finding the relationship between the eccentricities and solving for $\sin^2 \theta$, the length of the latus rectum of the ellipse can be calculated.

Option A is correct because the locus of the intersection of perpendicular tangents to a hyperbola is a circle with radius derived from the hyperbola's parameters.

Option B (1) is correct because the sum of the reciprocals of the squares of the eccentricities of a hyperbola and its conjugate is always equal to 1.

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

Option D (2) is correct because it involves finding the length of the latus rectum of a hyperbola given its foci and eccentricity.

Option C ($\alpha+\beta=\pi$) is correct because the sum of the angle subtended by the chord at the center and the angle between the tangents at the ends of the chord is equal to $\pi$.

Option A ($x = k, y = h$) is correct because, by rearranging the equation and shifting the origin, the asymptotes can be identified as x = k and y = h.

Option A ($8\sqrt 2$) is correct because it is the calculated distance between the directrices of the given hyperbola.

Option A ($\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{24}} = 1$) is correct because the equation satisfies the given vertices and directrix of the hyperbola.

Option D (Whose eccentricity is $\sqrt{\frac{19}{3}}$) is correct because after converting the equation to standard form, the eccentricity is calculated using the formula $e = \sqrt{1 + (b^2/a^2)}$.

Option C ($7(x-4)^2-9(y-3)^2=63$) is correct because it correctly uses the foci and eccentricity to derive the hyperbola equation.

Option B ($x^2+3 x y+2 y^2+2 x+3 y+2=0$) is correct because the conjugate hyperbola can be found by using the relationship H + C = 2A, where H is the original hyperbola, C is the conjugate hyperbola, and A is the pair of asymptotes.

Option C (Hyperbola) is correct because the equation can be rewritten in the form PS = ePM, where e = 3, which is greater than 1, indicating a hyperbola.