Hyperbola – JEE Main MathematicsPractice Questions & PYQs

The given conic, $\frac{x^2}{3} + \frac{y^2}{4} = 4$, can be rewritten in standard form by dividing by 4, which yields $\frac{x^2}{12} + \frac{y^2}{16} = 1$. This is an ellipse with its major axis along the $y$-axis, where $a^2 = 16$ and $b^2 = 12$. Its eccentricity is calculated as $e = \sqrt{1 - \frac{12}{16}} = \frac{1}{2}$, and the foci are located at $(0, \pm be) = (0, \pm 4 \cdot \frac{1}{2}) = (0, \pm 2)$.

Since the hyperbola has its vertices at the foci of this ellipse and its transverse axis along the major axis, it is a vertical hyperbola with the equation $\frac{y^2}{a_H^2} - \frac{x^2}{b_H^2} = 1$. The vertices at $(0, \pm 2)$ imply that $a_H = 2$. Given the eccentricity $e_H = \frac{3}{2}$, we use the relation $e_H^2 = 1 + \frac{b_H^2}{a_H^2}$ to find the value of $b_H^2$. Substituting the known values, we have $\frac{9}{4} = 1 + \frac{b_H^2}{4}$, which simplifies to $\frac{5}{4} = \frac{b_H^2}{4}$, confirming $b_H^2 = 5$.

The equation of the hyperbola is therefore $\frac{y^2}{4} - \frac{x^2}{5} = 1$. We can determine which point does not lie on the hyperbola by testing each option:
For $(0, 2)$, we have $\frac{2^2}{4} - \frac{0}{5} = 1 - 0 = 1$.
For $(\sqrt{5}, 2\sqrt{2})$, we have $\frac{(2\sqrt{2})^2}{4} - \frac{(\sqrt{5})^2}{5} = \frac{8}{4} - \frac{5}{5} = 2 - 1 = 1$.
For $(\sqrt{10}, 2\sqrt{3})$, we have $\frac{(2\sqrt{3})^2}{4} - \frac{(\sqrt{10})^2}{5} = \frac{12}{4} - \frac{10}{5} = 3 - 2 = 1$.
For $(\sqrt{5}, 2\sqrt{3})$, we have $\frac{(2\sqrt{3})^2}{4} - \frac{(\sqrt{5})^2}{5} = \frac{12}{4} - \frac{5}{5} = 3 - 1 = 2$.
Since $2 \neq 1$, the point $(\sqrt{5}, 2\sqrt{3})$ does not satisfy the equation of the hyperbola.

The geometry of a curve is determined by its normal line, which is perpendicular to the tangent at any point $P(x, y)$. Since the slope of the tangent is $\frac{dy}{dx}$, the slope of the normal is the negative reciprocal, $-\frac{dx}{dy}$, yielding the linear equation $Y - y = -\frac{dx}{dy}(X - x)$. Setting $Y = 0$ reveals the $x$-intercept $G$, where $X = x + y \frac{dy}{dx}$, placing the point $G$ at coordinates $(x + y \frac{dy}{dx}, 0)$.

The problem specifies that the distance from the origin to $G$ equals twice the abscissa of $P$, leading to the differential equation $x + y \frac{dy}{dx} = 2x$. This simplifies to $y \frac{dy}{dx} = x$, which can be integrated by separating the variables as $\int y \, dy = \int x \, dx$. Performing the integration results in $\frac{y^2}{2} = \frac{x^2}{2} + C$, which rearranges to $x^2 - y^2 = -2C$. This standard form confirms that the curve describes a hyperbola.

The position of the foci in a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ depends on the product of the semi-transverse axis $a$ and the eccentricity $e$. For the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$, identifying the axes reveals that $a = \cos \alpha$ and $b = \sin \alpha$.

Applying the standard relationship $b^2 = a^2(e^2 - 1)$ to determine the eccentricity leads to $\sin^2 \alpha = \cos^2 \alpha(e^2 - 1)$, which simplifies to $\tan^2 \alpha = e^2 - 1$ and therefore $e^2 = \sec^2 \alpha$, meaning $e = \sec \alpha$. Since the product $ae = \cos \alpha \cdot \sec \alpha = 1$, the x-coordinates of the foci are fixed at $\pm 1$ and do not vary with $\alpha$.

A line $y = mx + c$ serves as a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ only if it satisfies the condition $c^2 = a^2 m^2 - b^2$. By applying this to the line $y = \alpha x + \beta$, where the slope is $\alpha$ and the intercept is $\beta$, we obtain the relationship $\beta^2 = a^2 \alpha^2 - b^2$. Substituting the general coordinates $x$ and $y$ for the parameters $\alpha$ and $\beta$ transforms this into the equation
$a^2 x^2 - y^2 = b^2$
which reveals that the locus of the point $P(\alpha, \beta)$ is a hyperbola.

The foci of a conic section are located at a distance of $ae$ from the center along the major axis. Converting the hyperbola equation $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$ to standard form by multiplying the entire equation by $25$ results in $\frac{x^2}{(144/25)} - \frac{y^2}{(81/25)} = 1$. From this form, the semi-major axis is $a = \frac{12}{5}$ and the semi-minor axis is $b = \frac{9}{5}$, leading to an eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{81}{144}} = \frac{15}{12} = \frac{5}{4}$. Calculating the focal distance $ae$ gives $\frac{12}{5} \cdot \frac{5}{4} = 3$, meaning the foci are located at $(\pm 3, 0)$.

For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$, the semi-major axis is $a = 4$. Given that the foci are at the same positions, $ae = 3$, which implies $4e = 3$ or $e = \frac{3}{4}$. Applying the ellipse eccentricity relation $b^2 = a^2(1 - e^2)$ yields:
$b^2 = 16 \left( 1 - \left( \frac{3}{4} \right)^2 \right) = 16 \left( 1 - \frac{9}{16} \right) = 7$

A line $y = mx + c$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ when it satisfies the condition $c^2 = a^2m^2 - b^2$. The eccentricity $e = \sqrt{\frac{5}{2}}$ leads to the relationship $e^2 = 1 + \frac{b^2}{a^2} = \frac{5}{2}$, which simplifies to $\frac{b^2}{a^2} = \frac{3}{2}$, or $b^2 = \frac{3}{2}a^2$. The length of the latus rectum is expressed as $\frac{2b^2}{a} = 6\sqrt{2}$. Substituting the relationship $b^2 = \frac{3}{2}a^2$ into this formula gives $\frac{2(\frac{3}{2}a^2)}{a} = 6\sqrt{2}$, resulting in $3a = 6\sqrt{2}$ and solving to $a = 2\sqrt{2}$. With $a^2 = 8$, the value of $b^2$ becomes $12$.

Using the tangency condition for the line $y = 2x + c$, where the slope $m$ is 2, the equation becomes $c^2 = a^2m^2 - b^2$. Substituting the determined values $a^2 = 8$, $b^2 = 12$, and $m = 2$ provides the calculation $c^2 = 8(2)^2 - 12$. This simplifies to $32 - 12$, resulting in 20.

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the foci are given as $(\pm 5, 0)$, which corresponds to the coordinates $(\pm ae, 0)$, establishing that $ae = 5$. The standard relationship between the semi-axes and the eccentricity is $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$, which simplifies to $b^2 = a^2 - 25$. Given the length of the latus rectum as $\frac{2b^2}{a} = \sqrt{50} = 5\sqrt{2}$, we can substitute $b^2 = a^2 - 25$ into this expression to get $2(a^2 - 25) = 5\sqrt{2}a$. Rearranging this into the quadratic form $2a^2 - 5\sqrt{2}a - 50 = 0$ allows us to solve for $a$, resulting in $a = 5\sqrt{2}$. Plugging this back into our expression for $b^2$ gives $b^2 = (5\sqrt{2})^2 - 25 = 50 - 25 = 25$.

The given hyperbola equation is $\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1$, which is in the standard form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ where $A^2 = b^2$ and $B^2 = a^2b^2$. The square of the eccentricity for this hyperbola is defined by $e_H^2 = 1 + \frac{B^2}{A^2}$. Substituting the known values, we have $e_H^2 = 1 + \frac{a^2b^2}{b^2} = 1 + a^2$. Since $a^2 = (5\sqrt{2})^2 = 50$, the square of the eccentricity is $1 + 50 = 51$.

The standard equation of a hyperbola is determined by its vertices $(\pm 6, 0)$ and eccentricity $\frac{\sqrt{5}}{2}$, which establish the semi-major axis $a = 6$ and, using the relation $b^2 = a^2(e^2 - 1)$, the semi-minor axis $b = 3$. This results in the equation $\frac{x^2}{36} - \frac{y^2}{9} = 1$. The normal to this curve at any parametric point $(6 \sec \theta, 3 \tan \theta)$ is defined by the equation $\frac{6x}{\sec \theta} + \frac{3y}{\tan \theta} = 36 + 9$, which simplifies to $2x \cos \theta + y \cot \theta = 15$.

The geometric constraint imposed by the line $\sqrt{2} x + y = 2 \sqrt{2}$ requires the slope of the normal, $-2 \sin \theta$, to equal $-\sqrt{2}$. Solving $-2 \sin \theta = -\sqrt{2}$ gives $\theta = \frac{\pi}{4}$. Substituting this value results in the specific normal equation $\sqrt{2} x + y = 15$. Identifying the point $P(6 \sqrt{2}, 3)$ on the hyperbola and the intersection $K(0, 15)$ with the $y$-axis allows for the calculation of the squared distance $d^2 = (6 \sqrt{2} - 0)^2 + (3 - 15)^2 = 72 + 144 = 216$.

A hyperbola is uniquely determined by its semi-axes $a$ and $b$, which are related to its eccentricity $e$ by the identity $e^2 = 1 + \frac{b^2}{a^2}$. By determining these constants through the provided constraints, we can establish the dimensions of any hyperbola derived from them.

The eccentricity condition $15(e^2 + 1) = 34e$ can be rewritten as the quadratic equation $15e^2 - 34e + 15 = 0$. Factoring this leads to $(5e - 3)(3e - 5) = 0$, which gives $e = \frac{5}{3}$ since the eccentricity must be greater than 1. Using the identity $e^2 = 1 + \frac{b^2}{a^2}$, we find $\frac{b^2}{a^2} = (\frac{5}{3})^2 - 1 = \frac{16}{9}$, implying $b^2 = \frac{16a^2}{9}$. Substituting the point $(6, 4\sqrt{3})$ into the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ yields $\frac{36}{a^2} - \frac{48}{b^2} = 1$. Replacing $b^2$ with $\frac{16a^2}{9}$ results \in $\frac{36}{a^2} - \frac{48 × 9}{16a^2} = 1$, which simplifies to $\frac{36 - 27}{a^2} = 1$, establishing $a^2 = 9$ and $b^2 = 16$.

The second hyperbola equation is $\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$. With $a^2 = 9$ and $b^2 = 16$, this becomes $\frac{x^2}{16} - \frac{y^2}{2(9 + 1)} = 1$, or $\frac{x^2}{16} - \frac{y^2}{20} = 1$. For a hyperbola in the form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, the length of the latus rectum is given by $\frac{2B^2}{A}$. Here, $A = \sqrt{16} = 4$ and $B^2 = 20$. The final length is $\frac{2 \times 20}{4}$, which simplifies to 10.

Establishing the focal distance of a point on a hyperbola starts with determining the parameters of the associated ellipse. For the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$, the vertical orientation is apparent from the denominators, where $b^2=25$ and $a^2=9$. The eccentricity is $e = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$, which places the foci at $(0, \pm be) = (0, \pm 4)$. Since the hyperbola shares these foci, it follows a vertical structure $\frac{x^2}{A^2}-\frac{y^2}{B^2}=-1$. Given the hyperbola's eccentricity $e_H = \frac{4}{5} \times \frac{15}{8} = \frac{3}{2}$, the relation $B \cdot e_H = 4$ yields $B = \frac{8}{3}$, while the semi-transverse axis squared is $A^2 = B^2(e_H^2 - 1) = \frac{64}{9} \left( \frac{9}{4} - 1 \right) = \frac{80}{9}$.

The hyperbola's directrix is located at $y = \pm \frac{B}{e_H}$, which computes to $\pm \frac{16}{9}$. By definition, the focal distance of any point $P$ on a hyperbola is the product of the eccentricity and the distance from the point to the corresponding directrix, expressed as $PS = e_H \cdot PM$. Substituting the point $(\sqrt{2}, \frac{14}{3}\sqrt{\frac{2}{5}})$ and the directrix $y = \frac{16}{9}$ into this relationship results in the expression $\frac{3}{2} \left| \frac{14}{3}\sqrt{\frac{2}{5}} - \frac{16}{9} \right|$. Simplifying this yields $7\sqrt{\frac{2}{5}} - \frac{8}{3}$.

The condition for a line $y = mx + c$ to be tangent to the hyperbola $x^2 - y^2 = 1$ is $c^2 = m^2 - 1$, whereas for the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the condition is $c^2 = a^2 m^2 + b^2$. Because the line $y = \sqrt{\frac{5}{2}}x + K$ is a common tangent, we equate these two conditions to obtain $m^2 - 1 = a^2 m^2 + b^2$. Substituting the slope $m = \sqrt{\frac{5}{2}}$, this becomes $\frac{5}{2} - 1 = a^2(\frac{5}{2}) + b^2$, which simplifies to $5a^2 + 2b^2 = 3$.

The eccentricity of the hyperbola $x^2 - y^2 = 1$ is $\sqrt{2}$, making the eccentricity of the ellipse equal to $\frac{1}{\sqrt{2}}$. Applying the relationship $e^2 = 1 - \frac{b^2}{a^2}$ for the ellipse, we have $\frac{1}{2} = 1 - \frac{b^2}{a^2}$, which indicates that $a^2 = 2b^2$. Substituting $a^2 = 2b^2$ into our earlier equation yields $5(2b^2) + 2b^2 = 3$, simplifying to $12b^2 = 3$, or $b^2 = \frac{1}{4}$. It follows that $a^2 = \frac{1}{2}$, and the expression $4(a^2 + b^2)$ results in $4(\frac{1}{2} + \frac{1}{4}) = 3$.

The latus rectum of the hyperbola connects the points $(ae, \frac{b^2}{a})$ and $(ae, -\frac{b^2}{a})$ to the coordinate plane. Because the total angle subtended at the origin $(0,0)$ is $\frac{\pi}{3}$, the line segment connecting the origin to $(ae, \frac{b^2}{a})$ forms an angle of $\frac{\pi}{6}$ with the $x$-axis. Trigonometry dictates that the tangent of this half-angle is the ratio of the vertical distance to the horizontal distance, resulting in $\tan(\frac{\pi}{6}) = \frac{b^2/a}{ae} = \frac{b^2}{a^2 e}$. Substituting $a^2 = 9$ and $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$ leads to the relation $\frac{b^2}{9e} = \frac{1}{\sqrt{3}}$, which simplifies to $e = \frac{\sqrt{3}b^2}{9}$.

We invoke the standard hyperbola eccentricity relation $e^2 = 1 + \frac{b^2}{a^2}$, which becomes $e^2 = 1 + \frac{b^2}{9}$. Substituting our expression for $e$ into this equation gives $\frac{3b^4}{81} = 1 + \frac{b^2}{9}$. Multiplying throughout by $27$ simplifies this to the quadratic form $b^4 = 3b^2 + 27$, or $b^4 - 3b^2 - 27 = 0$. Solving for $b^2$ using the quadratic formula, we obtain $b^2 = \frac{3 \pm \sqrt{9 - 4(1)(-27)}}{2}$, which simplifies to $\frac{3}{2}(1 + \sqrt{13})$. Comparing this result to the given expression $\frac{l}{m}(1+\sqrt{n})$ allows us to identify $l=3$, $m=2$, and $n=13$. Calculating the required sum gives $l^2 + m^2 + n^2 = 3^2 + 2^2 + 13^2 = 182$.

The geometry of a hyperbola is fundamentally determined by the relationship between its foci and directrices. By setting the distance between the foci as $2ae = 6$ and the distance between the directrices as $\frac{2a}{e} = \frac{8}{3}$, we can solve for the primary parameters $a$ and $e$. Multiplying these equations yields $a^2 = 4$, while dividing them provides $e^2 = \frac{9}{4}$. Using the hyperbola identity $b^2 = a^2(e^2 - 1)$, we calculate $b^2 = 4(\frac{9}{4} - 1) = 5$, resulting in the equation $\frac{x^2}{4} - \frac{y^2}{5} = 1$.

When the vertical line $x = \alpha$ intersects this curve, the $y$-coordinates are found by rearranging the equation to $y^2 = \frac{5}{4}(\alpha^2 - 4)$, giving the points $A$ and $B$ as $(\alpha, \pm \frac{\sqrt{5}}{2}\sqrt{\alpha^2-4})$. The area of triangle $OAB$ with vertices at $(0,0)$, $A$, and $B$ is calculated as half the base along $x=\alpha$ multiplied by the height along the $x$-axis, resulting in $\frac{1}{2} \times \sqrt{5(\alpha^2 - 4)} \times |\alpha| = 4\sqrt{15}$. Squaring both sides produces $\frac{5}{4} \alpha^2 (\alpha^2 - 4) = 16 \times 15$, which simplifies to $\alpha^2(\alpha^2 - 4) = 192$. Setting $\alpha^2 = u$ allows us to solve $u^2 - 4u - 192 = 0$, where the positive value leads to $u = 16$.

The area of a triangle defined by a point on a hyperbola and its two foci is calculated as half the base, representing the distance between the foci, multiplied by the height, which is the ordinate of the point. For the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$, the parameters are $a^2 = 9$ and $b^2 = 4$, leading to an eccentricity calculated by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{9}} = \frac{\sqrt{13}}{3}$. Consequently, the distance between the foci is $2ae = 2 \cdot 3 \cdot \frac{\sqrt{13}}{3} = 2\sqrt{13}$.

Equating the triangle area of $2\sqrt{13}$ to the expression $\frac{1}{2} \cdot (2\sqrt{13}) \cdot y$ results in $y = 2$. Placing this value into the hyperbola equation gives $\frac{x^2}{9} - \frac{2^2}{4} = 1$, which simplifies to $\frac{x^2}{9} - 1 = 1$, indicating $x^2 = 18$. The square of the distance of the point from the origin is $x^2 + y^2$, so adding $18$ and $4$ yields $22$.

The geometric properties of a hyperbola define the relationship between its axes and eccentricity through the fundamental equation $e^2 = 1 + \frac{b^2}{a^2}$, where $a$ and $b$ represent the semi-transverse and semi-conjugate axis lengths. Given the eccentricity $e = \frac{\sqrt{11}}{2}$, we square this value to obtain $e^2 = \frac{11}{4}$, which simplifies the relationship to $1 + \frac{b^2}{a^2} = \frac{11}{4}$, or $\frac{b^2}{a^2} = \frac{7}{4}$. Taking the square root, we find the ratio $b = \frac{\sqrt{7}}{2}a$.

We are also told the sum of the full axis lengths $2a + 2b$ is $4(2\sqrt{2}+\sqrt{14})$, which implies $a + b = 4\sqrt{2} + 2\sqrt{14}$. Substituting the expression for $b$ into this sum gives $a(1 + \frac{\sqrt{7}}{2}) = 4\sqrt{2} + 2\sqrt{14}$. Factoring the right side as $2\sqrt{2}(2 + \sqrt{7})$ and the left side as $a(\frac{2+\sqrt{7}}{2})$, we cancel the common factor $(2 + \sqrt{7})$ to find $a = 4\sqrt{2}$, which results in $a^2 = 32$. Calculating $b$ from the established ratio yields $b = \frac{\sqrt{7}}{2}(4\sqrt{2}) = 2\sqrt{14}$, meaning $b^2 = 56$. Summing these values gives $a^2 + b^2 = 32 + 56 = 88$.

The geometry of the ellipse starts with the relationship between its eccentricity and the location of its directrices, expressed as $\frac{a}{e} = \frac{4\sqrt{6}}{3}$. Given $e = \frac{\sqrt{3}}{2}$, the semi-major axis $a$ evaluates to $2\sqrt{2}$. Using the standard ellipse identity $b^2 = a^2(1 - e^2)$, we find $b^2 = 8(1 - \frac{3}{4}) = 2$, identifying the length of the minor axis as $2b = 2\sqrt{2}$.

For the hyperbola, the conditions specify that its eccentricity $e_H$ matches the semi-major axis of the ellipse, $2\sqrt{2}$, while its latus rectum equals the ellipse's minor axis, $2\sqrt{2}$. Utilizing the hyperbola relations $B^2 = A^2(e_H^2 - 1)$ and the latus rectum length $\frac{2B^2}{A} = 2\sqrt{2}$, we substitute $e_H^2 = 8$ to find $B^2 = 7A^2$. Substituting this result into the latus rectum formula gives:

$\frac{2(7A^2)}{A} = 2\sqrt{2}$

This simplifies to $14A = 2\sqrt{2}$, yielding $A = \frac{\sqrt{2}}{7}$. The distance between the foci of the hyperbola is expressed as $2Ae_H$, which equals $2 \times \frac{\sqrt{2}}{7} \times 2\sqrt{2} = \frac{8}{7}$.

The hyperbola defined by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has foci at $(\pm ae, 0)$, where $e$ represents the eccentricity. The parabola shares the focus $(ae, 0)$ and has a directrix passing through the other focus at $(-ae, 0)$, which places the parabola's vertex at the origin and its focal length at $ae$. This configuration results in the standard parabola equation $y^2 = 4(ae)x$, which possesses a latus rectum of length $4ae$. Equating this to $e$ times the hyperbola's latus rectum length, $\frac{2b^2}{a}$, provides the relation $4ae = e \cdot \frac{2b^2}{a}$, which simplifies to $b^2 = 2a^2$.

Substituting the given point $(2\sqrt{2}, -2\sqrt{2})$ into the hyperbola's equation yields $\frac{8}{a^2} - \frac{8}{b^2} = 1$. By incorporating the relation $b^2 = 2a^2$ into this expression, the equation becomes $\frac{8}{a^2} - \frac{8}{2a^2} = 1$, which simplifies to $a^2 = 4$ and $b^2 = 8$. With $a=2$ and the eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 2} = \sqrt{3}$, the equation of the parabola is:

$y^2 = 8\sqrt{3}x$

Testing the provided options by substituting the coordinates into the parabola equation, the point $(3\sqrt{3}, -6\sqrt{2})$ yields $(-6\sqrt{2})^2 = 72$ for the left side and $8\sqrt{3} \cdot 3\sqrt{3} = 72$ for the right side. Since both sides are equal, this point lies on the parabola.

The eccentricity $e$ of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is related to its axes by the equation $b^2 = a^2(e^2 - 1)$. Given $e = \frac{5}{4}$, we find $b^2 = a^2 \left(\frac{25}{16} - 1\right) = a^2 \left(\frac{9}{16}\right)$, which implies $b = \frac{3}{4}a$.

Substituting the point $\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ into the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we get $\frac{64}{5 a^2} - \frac{144}{25 b^2} = 1$. Replacing $b^2$ with $\frac{9}{16}a^2$ yields $\frac{64}{5 a^2} - \frac{144}{25 \left(\frac{9}{16} a^2\right)} = 1$, which simplifies to $\frac{64}{5 a^2} - \frac{144 \cdot 16}{25 \cdot 9 a^2} = 1$. This becomes $\frac{64}{5 a^2} - \frac{256}{25 a^2} = 1$, or $\frac{320 - 256}{25 a^2} = 1$, leading to $a^2 = \frac{64}{25}$ and subsequently $b^2 = \frac{36}{25}$.

The equation of the normal to the hyperbola at point $(x_1, y_1)$ is given by $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$. Inserting $x_1 = \frac{8}{\sqrt{5}}$, $y_1 = \frac{12}{5}$, $a^2 = \frac{64}{25}$, and $b^2 = \frac{36}{25}$, we have $\frac{\frac{64}{25} x}{\frac{8}{\sqrt{5}}} + \frac{\frac{36}{25} y}{\frac{12}{5}} = \frac{64}{25} + \frac{36}{25}$. This simplifies to $\frac{8 \sqrt{5}}{25} x + \frac{3}{5} y = 4$. Multiplying the entire equation by $25$ results in $8 \sqrt{5} x + 15 y = 100$. Comparing this to the form $8 \sqrt{5} x + \beta y = \lambda$, we identify $\beta = 15$ and $\lambda = 100$. The value of $\lambda - \beta$ is $100 - 15 = 85$.