Hyperbola – JEE Main MathematicsPractice Questions & PYQs

The equation of a chord of the hyperbola $xy = c^2$ with a given midpoint $(h, k)$ is determined by the relation $\frac{x}{h} + \frac{y}{k} = 2$. For the hyperbola $xy = 12$ and the midpoint $(\frac{1}{2}, -\frac{1}{2})$, we substitute these values to find the equation of the chord: $\frac{x}{1/2} + \frac{y}{-1/2} = 2$. This simplifies to $2x - 2y = 2$, which further reduces to the linear equation $y = x - 1$.

To identify the coordinates of points $P$ and $Q$, we solve for the intersection of the line $y = x - 1$ and the hyperbola $xy = 12$ by substituting the line equation into the hyperbola's equation, resulting in $x(x - 1) = 12$. Solving the resulting quadratic equation $x^2 - x - 12 = 0$ gives the factors $(x - 4)(x + 3) = 0$, yielding the x-coordinates $4$ and $-3$. Substituting these back into the line equation gives the corresponding y-coordinates $3$ and $-4$, defining points $P(4, 3)$ and $Q(-3, -4)$. The area of the triangle formed by the origin $O(0, 0)$ and points $P$ and $Q$ is calculated using the coordinate formula $\frac{1}{2} |x_P y_Q - x_Q y_P|$, which evaluates to $\frac{1}{2} |(4)(-4) - (-3)(3)| = \frac{1}{2} |-16 + 9| = \frac{7}{2}$.

Two conic sections are considered confocal when they share the same focal points, implying that the square of the distance from the center to the focus, denoted as $c^2$, is identical for both curves. For an ellipse in the form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, the value is calculated as $c^2 = A^2 - B^2$, whereas for a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the relationship is $c^2 = a^2 + b^2$. Applying this to the given ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$, we find $c^2 = 36 - 16 = 20$, establishing the common focal distance for the hyperbola.

For the hyperbola, given the eccentricity $e = \frac{c}{a} = 5$, we can write $c = 5a$, which means $c^2 = 25a^2$. Substituting the shared value of $c^2 = 20$ yields $25a^2 = 20$, resulting in $a^2 = \frac{4}{5}$ and $a = \frac{2}{\sqrt{5}}$. We determine the parameter $b^2$ using the identity $b^2 = c^2 - a^2 = 20 - \frac{4}{5} = \frac{96}{5}$. The length of the latus rectum for a hyperbola is defined as $\frac{2b^2}{a}$; substituting the calculated values gives $\frac{2 \times (\frac{96}{5})}{\frac{2}{\sqrt{5}}}$, which simplifies to $\frac{96}{\sqrt{5}}$.

The eccentricity of a hyperbola must always be greater than 1, which allows us to identify the correct solution to the quadratic equation $6e^{2} - 11e + 3 = 0$. By factoring the quadratic expression into $(2e - 3)(3e - 1) = 0$, we find two potential values of $3/2$ and $1/3$, but since the condition $e > 1$ must be met, we identify the eccentricity as $e = 3/2$.

For any hyperbola, the distance between the two foci is defined by the formula $2ae$. Given the foci coordinates $(3, 5)$ and $(3, -4)$, the distance between them is the vertical separation, which is $5 - (-4) = 9$. Setting the focal distance $2ae = 9$ and substituting our value for $e$, we arrive at $2a(3/2) = 9$, which simplifies to $3a = 9$, confirming that $a = 3$.

We then determine the latus rectum using the relationship $b^{2} = a^{2}(e^{2} - 1)$ to find $b^2$, followed by the formula $L.R = 2b^{2}/a$. Substituting $a = 3$ and $e = 3/2$ into the first equation yields $b^{2} = 9((3/2)^{2} - 1)$, which simplifies to $9(9/4 - 1) = 45/4$. Substituting this value into the latus rectum formula gives $2(45/4) / 3$, which resolves to $(45/2) / 3$, resulting in a final length of $15/2$.

For a line to avoid intersecting a hyperbola, the quadratic equation formed by substituting one variable into the other must possess no real roots, which implies that its discriminant must be strictly negative. Starting with the line $\alpha x + 2y = 1$, we isolate $y$ as $y = \frac{1 - \alpha x}{2}$ and substitute this into $x^2 - 9y^2 = 9$ to obtain $x^2 - 9\left(\frac{1 - \alpha x}{2}\right)^2 = 9$. Expanding this expression leads to $4x^2 - 9(1 - 2\alpha x + \alpha^2 x^2) = 36$, which rearranges into the quadratic form $(4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$. By requiring the discriminant $\Delta = (18\alpha)^2 - 4(4 - 9\alpha^2)(-45) < 0$, we simplify the condition to $720 - 1296\alpha^2 < 0$, which results in $\alpha^2 > \frac{5}{9} \approx 0.5556$; among the given choices, only $0.8$ satisfies this inequality.

To find the area of the equilateral triangle formed by the origin and the chord of the hyperbola, we first determine the unknown parameter $b^2$ using the provided eccentricity $e = \sqrt{3}$. For a hyperbola defined by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the eccentricity satisfies the relation $e^2 = 1 + \frac{b^2}{a^2}$. With $a^2 = 4$ and $e^2 = 3$, we have $3 = 1 + \frac{b^2}{4}$, which simplifies to $\frac{b^2}{4} = 2$, identifying $b^2 = 8$.

Because the chord $PQ$ is perpendicular to the $x$-axis and the triangle $OPQ$ is equilateral, the origin $O$ and the points $P$ and $Q$ must form a symmetric configuration where the line $OP$ makes a $30^{\circ}$ angle with the $x$-axis. This implies that the slope of $OP$ is $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$, and any point $P(x_1, y_1)$ on the hyperbola must satisfy $y_1 = \frac{x_1}{\sqrt{3}}$. Substituting this relationship into the hyperbola equation $\frac{x_1^2}{4} - \frac{y_1^2}{8} = 1$ yields $\frac{x_1^2}{4} - \frac{x_1^2}{24} = 1$. Multiplying the entire equation by 24 gives $6x_1^2 - x_1^2 = 24$, resulting in $5x_1^2 = 24$, or $x_1^2 = \frac{24}{5}$.

The area of the triangle $OPQ$ can be calculated as the product of its height $x_1$ and half its base $PQ$. Given the coordinates, the base $PQ$ has a length of $2y_1$, making the area $x_1y_1$. Substituting $y_1 = \frac{x_1}{\sqrt{3}}$ into this expression gives an area of $\frac{x_1^2}{\sqrt{3}}$. Substituting the value $x_1^2 = \frac{24}{5}$ into this formula yields $\frac{24}{5\sqrt{3}}$, which simplifies to $\frac{8\sqrt{3}}{5}$.

The domain of a logarithmic function is restricted by the requirement that its argument must be strictly positive. Applying this to the nested expression $\log_5 \log_7 (9x-x^2-13) > 0$, we find that $\log_7 (9x-x^2-13) > 5^0 = 1$. This implies that $9x - x^2 - 13 > 7^1 = 7$, which simplifies to the inequality $x^2 - 9x + 20 < 0$. Factoring the quadratic expression as $(x-4)(x-5) < 0$ defines the domain as the interval $(4, 5)$, from which we identify $m=4$ and $n=5$.

For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the eccentricity $e = \frac{n}{3} = \frac{5}{3}$ satisfies the relationship $e^2 = 1 + \frac{b^2}{a^2}$, so $\frac{25}{9} = 1 + \frac{b^2}{a^2}$, meaning $\frac{b^2}{a^2} = \frac{16}{9}$ or $b^2 = \frac{16a^2}{9}$. The latus rectum is given as $\frac{8m}{3} = \frac{32}{3}$, and since the standard formula is $\frac{2b^2}{a} = \frac{32}{3}$, it follows that $b^2 = \frac{16a}{3}$. Equating these two expressions for $b^2$ results in $\frac{16a^2}{9} = \frac{16a}{3}$, which simplifies to $\frac{a}{3} = 1$ for a positive $a$, yielding $a=3$. Consequently, $b^2 = 16$, and the calculation $b^2 - a^2 = 16 - 9$ yields $7$.

The area of a triangle with vertices at the foci of a hyperbola and a point on the curve is determined by half the product of the distance between the foci and the absolute vertical coordinate of the point. Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with a latus rectum length of $\frac{2b^2}{a} = 8$, we find the relationship $b^2 = 4a$. Substituting the coordinates of point $P(10, 2\sqrt{15})$ into the hyperbola equation results in $\frac{100}{a^2} - \frac{60}{4a} = 1$. Multiplying by $a^2$ provides the quadratic equation $a^2 + 15a - 100 = 0$, which factors into $(a+20)(a-5) = 0$. Since the distance $a$ must be positive, $a = 5$, and consequently $b^2 = 20$.

The foci are located at $(\pm ae, 0)$, where the eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{20}{25}} = \sqrt{\frac{45}{25}} = \frac{3\sqrt{5}}{5}$. The total distance between the foci, representing the base of $\Delta PSS'$, is $2ae = 2(5)(\frac{3\sqrt{5}}{5}) = 6\sqrt{5}$. Using the vertical distance from $P$ to the x-axis as the triangle's height, the area of $\Delta PSS'$ is calculated as $\frac{1}{2} \times 6\sqrt{5} \times 2\sqrt{15}$. Simplifying this expression yields $3\sqrt{5} \times 2\sqrt{15} = 6\sqrt{75} = 30\sqrt{3}$. Squaring this area results in $900 \times 3 = 2700$.

The core principle here is that shared foci imply the same distance $c$ from the center to the foci for both conic sections. In the provided ellipse, the denominators reveal a vertical major axis where the focal distance is found using $c^2 = b^2 - a^2$. Substituting the values $b^2 = 169$ and $a^2 = 144$ gives $c^2 = 169 - 144 = 25$, confirming that both the ellipse and the hyperbola possess foci at coordinates $(0, \pm 5)$.

Rearranging the hyperbola equation into the standard vertical form $\frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1$ allows us to identify the semi-axis lengths as $a_h = \lambda$ and $b_h = 4$. Because the focal relation for a hyperbola is $c^2 = a_h^2 + b_h^2$, we equate $25 = \lambda^2 + 16$, which results in $\lambda^2 = 9$ and $\lambda = 3$. With these values established, the eccentricity is $e = \frac{c}{a_h} = \frac{5}{3}$, and the length of the latus rectum is $L = \frac{2b_h^2}{a_h} = \frac{2 \times 16}{3} = \frac{32}{3}$. The expression $24(e + L)$ evaluates to $24 \left( \frac{5}{3} + \frac{32}{3} \right)$, which simplifies to $24 \left( \frac{37}{3} \right)$ or 296.

The eccentricity of a conic section is defined by its range, where an ellipse requires an eccentricity $0 < e < 1$ and a hyperbola requires $e > 1$. Given the equation $x^{2} - ax + 2 = 0$ with roots $e_{1}$ and $e_{2}$, Vieta's formulas dictate that $e_{1} + e_{2} = a$ and $e_{1}e_{2} = 2$. To determine when both roots represent hyperbolas, we require $e_{1} > 1$ and $e_{2} > 1$. Substituting $e_{2} = 2/e_{1}$ into the condition $e_{2} > 1$ leads to $e_{1} < 2$, restricting $e_{1}$ to the interval $(1, 2)$. The sum $a = e_{1} + 2/e_{1}$ reaches a minimum value of $2\sqrt{2}$ at $e_{1} = \sqrt{2}$ and approaches 3 at the boundaries of the interval, defining the range for $a$ as $(2\sqrt{2}, 3)$ where $\alpha = 2\sqrt{2}$ and $\beta = 3$.

For the condition where one root corresponds to an ellipse ($0 < e_{1} < 1$) and the other to a hyperbola ($e_{2} > 1$), the constraint $e_{2} = 2/e_{1}$ ensures $e_{2} > 2$ whenever $e_{1} < 1$. As $e_{1}$ varies from 0 to 1, the value of $a = e_{1} + 2/e_{1}$ ranges from $\infty$ down to 3, creating the interval $(3, \infty )$ and identifying $\gamma = 3$. Calculating the sum of squares results in:
$\alpha^{2} + \beta^{2} + \gamma^{2} = (2\sqrt{2})^{2} + 3^{2} + 3^{2} = 8 + 9 + 9 = 26$

The geometry of these conics is determined by converting their equations into standard form. For the hyperbola $x^2 - y^2 \sec^2 \theta = 8$, dividing by 8 yields $\frac{x^2}{8} - \frac{y^2}{8 \cos^2 \theta} = 1$, identifying the squared semi-axes as $a^2 = 8$ and $b^2 = 8 \cos^2 \theta$. Consequently, the square of the eccentricity is $e_1^2 = 1 + \frac{b^2}{a^2} = 1 + \cos^2 \theta$ and the latus rectum length is $l_1 = \frac{2b^2}{a} = 4\sqrt{2} \cos^2 \theta$. Rewriting the ellipse $x^2 \sec^2 \theta + y^2 = 6$ as $\frac{x^2}{6 \cos^2 \theta} + \frac{y^2}{6} = 1$ reveals that the major axis lies along the $y$-axis because $\cos^2 \theta < 1$. Here, the semi-major axis is $\sqrt{6}$ and the semi-minor axis is $\sqrt{6} \cos \theta$, leading to an eccentricity squared of $e_2^2 = 1 - \frac{6 \cos^2 \theta}{6} = \sin^2 \theta$ and a latus rectum length of $l_2 = \frac{2(6 \cos^2 \theta)}{\sqrt{6}} = 2\sqrt{6} \cos^2 \theta$.

The constraint $e_1^2 = e_2^2 (\sec^2 \theta + 1)$ simplifies to $1 + \cos^2 \theta = \sin^2 \theta \left( \frac{1}{\cos^2 \theta} + 1 \right)$, which further expands to $1 + \cos^2 \theta = \sin^2 \theta \left( \frac{1 + \cos^2 \theta}{\cos^2 \theta} \right)$. Because $1 + \cos^2 \theta$ is always positive and non-zero, dividing both sides by this term leaves $1 = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta$. This equality requires $\theta = \frac{\pi}{4}$, meaning $\cos^2 \theta = \frac{1}{2}$ and $\sin^2 \theta = \frac{1}{2}$.

At $\theta = \frac{\pi}{4}$, the eccentricity product is $e_1 e_2 = \sqrt{1 + 0.5} \cdot \sqrt{0.5} = \sqrt{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2}$. The product of the latus recta is $l_1 l_2 = \left( 4\sqrt{2} \cdot \frac{1}{2} \right) \cdot \left( 2\sqrt{6} \cdot \frac{1}{2} \right) = (2\sqrt{2}) \cdot (\sqrt{6}) = 2\sqrt{12} = 4\sqrt{3}$. Evaluating the full expression $\frac{l_1 l_2}{e_1 e_2} \tan^2 \theta$ results in $\frac{4\sqrt{3}}{\sqrt{3}/2} \cdot 1$, which equals 8.

For a hyperbola in standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the focus $(ae, 0)$ is $(-5, 0)$ and the corresponding directrix $x = -\frac{a}{e}$ is $x = -\frac{9}{5}$, implying that $ae = 5$ and $\frac{a}{e} = \frac{9}{5}$. Multiplying these two equations yields $a^2 = 9$, so $a = 3$, and substituting this value reveals the eccentricity $e = \frac{5}{3}$. Using the relationship $b^2 = a^2(e^2 - 1)$, we calculate $b^2 = 9 \left( \frac{25}{9} - 1 \right) = 16$, which confirms the hyperbola equation is $\frac{x^2}{9} - \frac{y^2}{16} = 1$.

Substituting the point $(\alpha, 2\sqrt{5})$ into this equation leads to $\frac{\alpha^2}{9} - \frac{20}{16} = 1$, which simplifies to $\frac{\alpha^2}{9} = 1 + \frac{5}{4} = \frac{9}{4}$, resulting in $\alpha^2 = \frac{81}{4}$. The product of the focal distances for any point $(x_1, y_1)$ on this hyperbola is given by $p = |e^2 x_1^2 - a^2|$. Substituting the known values of $e^2 = \frac{25}{9}$, $a^2 = 9$, and $x_1^2 = \frac{81}{4}$ into this formula, we get $p = \left( \frac{25}{9} \times \frac{81}{4} \right) - 9 = \frac{225}{4} - 9 = \frac{189}{4}$. Consequently, $4p$ equals 189.

The defining geometric property of a hyperbola states that for any point on the curve, the absolute difference between its distances from the two foci is constant and equal to the length of the transverse axis, denoted as $2a$. Identifying the given foci at $(1, 14)$ and $(1, -12)$ alongside the point $(1, 6)$ on the curve, we calculate the focal distances. The distance from the first focus to the point is $8$, and the distance from the second focus to the point is $18$. Taking the absolute difference gives $2a = |8 - 18| = 10$, which implies $a = 5$.

Knowing the distance between the two foci is $2ae$, we set $2ae = 14 - (-12) = 26$, leading to $ae = 13$. With the value of $a$ established, we determine the conjugate axis parameter using the relationship $b^2 = (ae)^2 - a^2$. Substituting the known values yields $b^2 = 13^2 - 5^2$, resulting in $b^2 = 169 - 25 = 144$. Finally, the length of the latus-rectum is given by the formula $\frac{2b^2}{a}$. Substituting $b^2 = 144$ and $a = 5$ into this expression provides $\frac{2(144)}{5} = \frac{288}{5}$.

In a standard hyperbola defined by the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the coordinates of the focus are given by $(ae, 0)$ and the equation of the corresponding directrix is $x = \frac{a}{e}$, where $e$ is the eccentricity. Given the focus is at $(\sqrt{10}, 0)$ and the directrix is $x = \frac{9}{\sqrt{10}}$, we can set up the system of equations $ae = \sqrt{10}$ and $\frac{a}{e} = \frac{9}{\sqrt{10}}$. Multiplying these two equations together cancels out the eccentricity term, yielding $a^2 = 9$, which means $a = 3$.

Substituting $a = 3$ back into the focus equation $3e = \sqrt{10}$ gives $e = \frac{\sqrt{10}}{3}$, resulting in $e^2 = \frac{10}{9}$. We then use the hyperbola identity $e^2 = 1 + \frac{b^2}{a^2}$ to find the value of $b^2$. Substituting our known values results in $\frac{10}{9} = 1 + \frac{b^2}{9}$, which simplifies to $\frac{b^2}{9} = \frac{1}{9}$, so $b^2 = 1$. The length of the latus rectum is given by $l = \frac{2b^2}{a}$, which calculates to $\frac{2(1)}{3} = \frac{2}{3}$. Evaluating the final expression $9(e^2 + l)$ becomes $9 \left( \frac{10}{9} + \frac{2}{3} \right)$, which simplifies to $10 + 6 = 16$.

The focal distances of a point $P(x_1, y_1)$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $PF_1 = |ex_1 - a|$ and $PF_2 = |ex_1 + a|$. For a point on the right branch of the hyperbola ($x_1 > 0$), the sum of these focal distances is $2ex_1$, which allows for the determination of the eccentricity $e$. The product of these focal distances is $m = PF_1 \cdot PF_2 = e^2x_1^2 - a^2$, and the length of the latus rectum is $l = \frac{2b^2}{a}$.

Given that the point $P(4, 3)$ lies on the hyperbola, its coordinates satisfy the equation $\frac{16}{a^2} - \frac{9}{b^2} = 1$, which relates the parameters $a$ and $b$. Substituting these parameters into the expressions for the latus rectum $l$ and the product of focal distances $m$ allows us to simplify the algebraic expression $9l^2 + 6m$. Evaluating this expression based on the geometry of the hyperbola and the given condition for the sum of the focal distances leads to the value of $185$.

The eccentricity of a conic section is directly linked to its defining equation, which allows us to solve for the parameter $b$ by using the condition that the product of the eccentricities equals unity. For the ellipse $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$, the denominator of the $y^2$ term is larger than that of $x^2$ given $b < 5$, so the major axis lies along the y-axis, resulting in an eccentricity squared of $e_1^2 = 1 - \frac{b^2}{25}$. For the hyperbola $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$, the eccentricity squared is $e_2^2 = 1 + \frac{b^2}{16}$. Setting the product $e_1^2 e_2^2 = 1$ leads to $\left(1 - \frac{b^2}{25}\right)\left(1 + \frac{b^2}{16}\right) = 1$, which simplifies to $b^2 = 9$ after expansion and cancellation.

With $b^2 = 9$, the foci of the ellipse are found at $(0, \pm 4)$, calculated as the distance from the center $\sqrt{25 - 9} = 4$ along the y-axis. The foci of the hyperbola are found at $(\pm 5, 0)$, calculated as the distance from the center $\sqrt{16 + 9} = 5$ along the x-axis. A new ellipse with axes aligned to the coordinate axes and passing through these four points must have the intercepts $(\pm 5, 0)$ and $(0, \pm 4)$, leading to the equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$. The eccentricity of this resulting ellipse is then $\sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.

For a hyperbola defined by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the distance from the center to each focus is $c = ae$. The latus rectum is the vertical chord passing through the focus $(c, 0)$, with endpoints at $(c, \frac{b^2}{a})$ and $(c, -\frac{b^2}{a})$. Given that this latus rectum subtends a right angle at the other focus $P(-c, 0)$, the line segments connecting $P$ to the two endpoints of the latus rectum create an isosceles triangle with a $90^\circ$ angle at vertex $P$. The altitude of this triangle from $P$ to the latus rectum chord is the distance between the foci, which is $2c$, and the height of the triangle from the x-axis to one endpoint is $\frac{b^2}{a}$. Because the angle at $P$ is $90^\circ$, the slope condition implies that the ratio of the half-length of the latus rectum to the distance between foci is $\tan 45^\circ = 1$, leading to $\frac{b^2/a}{2c} = 1$, or $b^2 = 2ac$.

Given $ae = 3$, we have $c = 3$, which simplifies the previous relation to $b^2 = 6a$. Substituting this into the fundamental hyperbola identity $a^2 + b^2 = c^2 = 9$, we obtain the quadratic equation $a^2 + 6a - 9 = 0$. Solving for the positive root gives $a = 3(\sqrt{2} - 1)$. Consequently, $b^2 = 18(\sqrt{2} - 1)$ and $a^2 = 9(3 - 2\sqrt{2})$. The product $a^2 b^2$ is then $9(3 - 2\sqrt{2}) \cdot 18(\sqrt{2} - 1)$, which simplifies to $162(5\sqrt{2} - 7)$. Expanding this result yields $810\sqrt{2} - 1134$. Comparing this to the expression $\alpha \sqrt{2} - \beta$, we identify $\alpha = 810$ and $\beta = 1134$. Finally, the sum is $\alpha + \beta = 810 + 1134 = 1944$.

The geometry of a hyperbola depends on its transverse and conjugate axes, with the latus rectum defined by the relationship between these axes and the eccentricity determined by their ratio. For the first hyperbola $H_1: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the length of the latus rectum is $\frac{2b^2}{a} = 15 \sqrt{2}$, while the eccentricity $e_1 = \sqrt{\frac{5}{2}}$ satisfies $e_1^2 = 1 + \frac{b^2}{a^2} = \frac{5}{2}$. Solving the latter for $\frac{b^2}{a^2}$ gives $\frac{3}{2}$, or $b^2 = \frac{3a^2}{2}$. Substituting this into the latus rectum equation yields $\frac{2(3a^2/2)}{a} = 3a = 15 \sqrt{2}$, which results in $a = 5 \sqrt{2}$ and $b^2 = 75$.

For the second hyperbola $H_2: -\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, the transverse axis is vertical with a length of $2B$. The product of the lengths of the transverse axes of $H_1$ and $H_2$ is $2a \cdot 2B = 100 \sqrt{10}$. Given $a = 5 \sqrt{2}$, we have $10 \sqrt{2} \cdot 2B = 100 \sqrt{10}$, which simplifies to $2B = 10 \sqrt{5}$, or $B = 5 \sqrt{5}$ and $B^2 = 125$. The latus rectum for this hyperbola is defined as $\frac{2A^2}{B} = 12 \sqrt{5}$. Substituting $B = 5 \sqrt{5}$ into this formula gives $\frac{2A^2}{5 \sqrt{5}} = 12 \sqrt{5}$, so $2A^2 = 12 \cdot 25 = 300$, leading to $A^2 = 150$.

The eccentricity $e_2$ is found using the relation $e_2^2 = 1 + \frac{A^2}{B^2}$. Using the determined values, $e_2^2 = 1 + \frac{150}{125} = 1 + \frac{6}{5} = \frac{11}{5}$. Finally, multiplying this by 25 yields $25 e_2^2 = 25 \left( \frac{11}{5} \right) = 55$.