Hyperbola – JEE Main MathematicsPractice Questions & PYQs

The length of the latus rectum for a conic section provides a direct relationship between its geometric parameters. For the ellipse $3x^2+4y^2=12$, which simplifies to the standard form $\frac{x^2}{4} + \frac{y^2}{3} = 1$, the length of the latus rectum is calculated as $\frac{2b^2}{a} = \frac{2(3)}{2} = 3$. Given that the hyperbola $\frac{x^2}{a^2} - y^2 = 1$ has the same latus rectum length, we equate $\frac{2(1)^2}{a} = 3$, which yields the value $a = \frac{2}{3}$.

Determining the squared eccentricities follows the specific formulas for each curve, with $e_H^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{1}{(2/3)^2} = 1 + \frac{9}{4}$ for the hyperbola and $e_E^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{3}{4}$ for the ellipse. Summing these values gives $e_H^2 + e_E^2 = \frac{13}{4} + \frac{1}{4} = \frac{14}{4}$. Multiplying this sum by 12 results in $12 \times \frac{14}{4} = 3 \times 14 = 42$.

The location of the foci for a central conic depends on the semi-major axis $a$ and eccentricity $e$, where the focal distance from the center is $c = ae$. For the ellipse $\frac{x^2}{16} + \frac{y^2}{7} = 1$, the semi-major axis $a$ is 4 and the eccentricity $e = \sqrt{1 - \frac{7}{16}} = \frac{3}{4}$, identifying the foci at $(\pm 3, 0)$. The hyperbola $\frac{x^2}{144} - \frac{y^2}{\alpha} = \frac{1}{25}$ can be written in standard form as $\frac{x^2}{(12/5)^2} - \frac{y^2}{(\sqrt{\alpha}/5)^2} = 1$, requiring its focal distance to satisfy the relation $c^2 = a^2 + b^2$. Substituting $c=3$ and $a = \frac{12}{5}$ gives $9 = \frac{144}{25} + \frac{\alpha}{25}$, which solves to $\alpha = 81$. Using the standard formula for the length of the latus rectum, $\frac{2b^2}{a}$, the calculation $\frac{2 \cdot (81/25)}{12/5}$ results in $\frac{27}{10}$.

Finding the coordinates of the tangency point is the initial step, as the slope of the tangent to the parabola $y^2=24x$ at $(\alpha, \beta)$ is determined by the expression $\beta y=12(x+\alpha)$, which simplifies to a slope of $\frac{12}{\beta}$. Because the tangent must be perpendicular to the line $2x+2y=5$, which has a slope of $-1$, the tangent slope must be $1$. Equating $\frac{12}{\beta}=1$ gives $\beta=12$, and substituting this coordinate into the parabola equation reveals $\alpha=6$.

With the hyperbola equation defined as $\frac{x^2}{36}-\frac{y^2}{144}=1$, the normal is constructed at the point $(\alpha+4, \beta+4)$, which evaluates to $(10, 16)$. Applying the formula for the normal to a hyperbola at a specific point, the equation $\frac{36x}{10}+\frac{144y}{16}=36+144$ represents the line. Simplifying this yields the linear form $\frac{x}{50}+\frac{y}{20}=1$. Checking the provided coordinate options against this equation confirms that the line does not pass through $(15, 13)$, as substituting these values results in $\frac{15}{50}+\frac{13}{20} = 0.3+0.65 = 0.95$, which does not equal $1$.

For a hyperbola defined by $k x^2 - y^2 = 6$, we first express it in standard form as $\frac{x^2}{6/k} - \frac{y^2}{6} = 1$. In this configuration, the distance from the center to the directrix is given by $\frac{a}{e} = 1$, which implies $a^2 = e^2$. Since the square of the eccentricity is defined as $e^2 = 1 + \frac{b^2}{a^2}$, we can substitute the known values $a^2 = \frac{6}{k}$ and $b^2 = 6$ to find $e^2 = 1 + \frac{6}{6/k} = 1 + k$. Equating $a^2$ and $e^2$ results in $\frac{6}{k} = 1 + k$, which simplifies to the quadratic equation $k^2 + k - 6 = 0$. Factoring this expression as $(k+3)(k-2) = 0$ gives $k=2$ as the required positive constant. With $k=2$, the hyperbola equation becomes $2x^2 - y^2 = 6$. Testing the point $(\sqrt{5}, -2)$ in this equation confirms that $2(\sqrt{5})^2 - (-2)^2 = 10 - 4 = 6$, showing that the hyperbola passes through this coordinate.

The eccentricity $e$ of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ relates to its semi-axes through the identity $e^{2}=1+\frac{b^{2}}{a^{2}}$. Given $e=\frac{\sqrt{5}}{2}$, we have $e^{2}=\frac{5}{4}$, which implies $1+\frac{b^{2}}{a^{2}}=\frac{5}{4}$, simplifying to $\frac{b^{2}}{a^{2}}=\frac{1}{4}$ or $a^{2}=4b^{2}$.

Since the point $P(-2\sqrt{6},\sqrt{3})$ lies on the hyperbola, its coordinates satisfy the equation $\frac{(-2\sqrt{6})^{2}}{a^{2}}-\frac{(\sqrt{3})^{2}}{b^{2}}=1$. Substituting $a^{2}=4b^{2}$ into this expression results in $\frac{24}{4b^{2}}-\frac{3}{b^{2}}=1$, which simplifies to $\frac{6}{b^{2}}-\frac{3}{b^{2}}=1$, leading to $b^{2}=3$ and $a^{2}=12$.

The tangent at $P(x_1, y_1)$ is defined by the equation $\frac{x x_1}{a^{2}}-\frac{y y_1}{b^{2}}=1$. Substituting the known values yields $\frac{x(-2\sqrt{6})}{12}-\frac{y(\sqrt{3})}{3}=1$. To find the intersection $Q$ with the conjugate axis, we set $x=0$, which gives $-\frac{y}{\sqrt{3}}=1$, identifying point $Q$ as $(0,-\sqrt{3})$.

For the normal at $P$, the slope of the tangent is calculated as $-\frac{x_1/a^2}{y_1/b^2} = -\frac{-2\sqrt{6}/12}{\sqrt{3}/3} = -\frac{\sqrt{2}}{2}$, meaning the normal slope is the negative reciprocal, $\sqrt{2}$. Using the point-slope formula $y-\sqrt{3}=\sqrt{2}(x+2\sqrt{6})$, we determine the intersection $R$ with the $y$-axis by setting $x=0$, resulting in $y=4\sqrt{3}+\sqrt{3}=5\sqrt{3}$, so point $R$ is $(0,5\sqrt{3})$. The length $QR$ is the absolute difference between the $y$-coordinates of $R$ and $Q$, which is $5\sqrt{3}-(-\sqrt{3})=6\sqrt{3}$.

The task of finding the locus of the intersection point of two lines involving a variable parameter requires us to eliminate that parameter to reveal the geometric relationship between the coordinates $x$ and $y$. By treating the intersection point $(x, y)$ as satisfying both equations $\sqrt{3} k x + k y - 4\sqrt{3}=0$ and $\sqrt{3} x - y - 4\sqrt{3} k=0$, we can manipulate the system algebraically to solve for the parameter $k$.

Through standard elimination methods, such as solving the linear system for $x$ and $y$, we find that the coordinates of the intersection point can be expressed as $x = 2(k + \frac{1}{k})$ and $y = 2\sqrt{3}(\frac{1}{k} - k)$. Dividing these expressions isolates the parameter terms, yielding $\frac{x}{2} = k + \frac{1}{k}$ and $\frac{y}{2\sqrt{3}} = \frac{1}{k} - k$. To eliminate $k$ and uncover the path of the locus, we square both equations and compute their difference. This calculation results in:
$\frac{x^2}{4} - \frac{y^2}{12} = \left(k + \frac{1}{k}\right)^2 - \left(\frac{1}{k} - k\right)^2 = 4$
Simplifying this expression leads to the standard form of a hyperbola, $\frac{x^2}{16} - \frac{y^2}{48} = 1$. Identifying the semi-axes from this equation gives $a^2 = 16$ and $b^2 = 48$. Using the relationship for eccentricity, $e = \sqrt{1 + \frac{b^2}{a^2}}$, we substitute the values to obtain $e = \sqrt{1 + \frac{48}{16}}$. This simplifies to $\sqrt{1 + 3}$, which results in an eccentricity of 2.

The centroid of a triangle is the arithmetic mean of the coordinates of its vertices, which directly links the locus of the centroid to any point on the hyperbola. The given equation $16x^2 - 9y^2 + 32x + 36y - 164 = 0$ can be rewritten by completing the squares as $16(x+1)^2 - 9(y-2)^2 = 144$, which describes a hyperbola centered at $(-1, 2)$ with foci at $(4, 2)$ and $(-6, 2)$. If we denote the centroid as $(x, y)$ and a variable point on the hyperbola as $(\alpha, \beta)$, the definitions for the centroid yield $x = \frac{\alpha - 6 + 4}{3}$ and $y = \frac{\beta + 2 + 2}{3}$, leading to the relations $\alpha = 3x + 2$ and $\beta = 3y - 4$. Substituting these expressions for $\alpha$ and $\beta$ into the hyperbola equation $16(\alpha+1)^2 - 9(\beta-2)^2 = 144$ results in $16(3x+3)^2 - 9(3y-6)^2 = 144$, which simplifies upon division by $9$ to $16(x+1)^2 - 9(y-2)^2 = 16$. Expanding this equation yields $16x^2 - 9y^2 + 32x + 36y - 36 = 0$.

The equation of a chord of the circle $x^2 + y^2 = 25$ having a midpoint at $(h, k)$ is determined by the formula $T = S_1$, which simplifies to $hx + ky = h^2 + k^2$. Expressing this line in the slope-intercept form $y = mx + c$ yields $y = -\frac{h}{k}x + \frac{h^2 + k^2}{k}$, identifying the slope as $m = -\frac{h}{k}$ and the y-intercept as $c = \frac{h^2 + k^2}{k}$.

A line $y = mx + c$ serves as a tangent to the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$ when it satisfies the condition $c^2 = a^2m^2 - b^2$, which corresponds to $c^2 = 9m^2 - 16$. Substituting the slope and intercept derived from the chord equation into this tangency condition results in $\left(\frac{h^2 + k^2}{k}\right)^2 = 9\left(-\frac{h}{k}\right)^2 - 16$. Multiplying both sides by $k^2$ to eliminate the denominators leads to the equation $(h^2 + k^2)^2 = 9h^2 - 16k^2$. Replacing the coordinates $(h, k)$ with general variables $(x, y)$ provides the final locus as $(x^2 + y^2)^2 - 9x^2 + 16y^2 = 0$.

The geometry of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{2} = 1$ is defined by semi-axes $a=2$ and $b=\sqrt{2}$, resulting in an eccentricity $e = \sqrt{\frac{3}{2}}$ and a focus $F(\sqrt{6}, 0)$ nearest to the point $P(4, \sqrt{6})$. Applying the tangent formula at $P$ provides the line $2x - \sqrt{6}y = 2$, which intersects the x-axis at $Q(1, 0)$ and meets the latus rectum $x = \sqrt{6}$ at $R\left(\sqrt{6}, \frac{2(\sqrt{6}-1)}{\sqrt{6}}\right)$. Because $Q$ and $F$ lie on the x-axis with a separation of $\sqrt{6}-1$, the area of $\triangle QFR$ is calculated as $\frac{1}{2} \times (\sqrt{6}-1) \times \frac{2(\sqrt{6}-1)}{\sqrt{6}}$, which simplifies to $\frac{7}{\sqrt{6}} - 2$.

The chord of the hyperbola $x^{2}-y^{2}=4$ with a midpoint at $(h, k)$ is described by the equation $T=S_{1}$, which simplifies to $xh-yk=h^{2}-k^{2}$. Because this chord also functions as a tangent to the parabola $y^{2}=8x$, it must be equivalent to the standard tangent form $y=mx+\frac{2}{m}$, which rearranges into the linear form $m^{2}x-my+2=0$.

Comparing the coefficients of these two linear equations requires that the ratios $\frac{h}{m^{2}}=\frac{-k}{-m}=\frac{h^{2}-k^{2}}{-2}$ hold true. Solving the first ratio gives the slope $m=\frac{h}{k}$, and substituting this into the equality $\frac{k}{m}=\frac{h^{2}-k^{2}}{-2}$ results in the expression $\frac{k^{2}}{h}=\frac{h^{2}-k^{2}}{-2}$. Rearranging this equation leads to $-2k^{2}=h^{3}-hk^{2}$, which simplifies further to $h^{3}=k^{2}(h-2)$, confirming the locus of the mid-points is $x^{3}=y^{2}(x-2)$.

A line tangent to a conic section must satisfy specific geometric conditions linking its slope and intercept to the parameters of that conic. For the line $2x+y=k$, rewriting it as $y=-2x+k$ reveals a slope of $m=-2$. For a hyperbola defined by $x^2-y^2=3$, the condition for a line $y=mx+c$ to be tangent is $c^2 = a^2m^2 - b^2$, where $a^2=3$ and $b^2=3$. Substituting the slope and hyperbola parameters results in $k^2 = 3(-2)^2 - 3$, which simplifies to $k^2=9$, and since $k>0$, we determine that $k=3$.

Now that the line is fixed as $y=-2x+3$, we apply the condition for tangency to the parabola $y^2=\alpha x$. The general form for a line tangent to this parabola is $y=mx+\frac{\alpha}{4m}$. Equating the constant term $3$ to $\frac{\alpha}{4(-2)}$ allows us to solve for the unknown, giving $3 = -\frac{\alpha}{8}$, which results in $\alpha = -24$.

The hyperbola $2x^2 - y^2 = 2$ is written in standard form as $x^2 - \frac{y^2}{2} = 1$, identifying parameters $a^2 = 1$ and $b^2 = 2$. For any point $(x_1, y_1)$ on a hyperbola, the normal line is given by the equation $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$. Substituting the coordinates of point $A(\sec \theta, 2 \tan \theta)$, the normal equation becomes $\frac{x}{\sec \theta} + \frac{2y}{2 \tan \theta} = 1 + 2$, which simplifies to $x + y \csc \theta = 3 \sec \theta$.

Applying the condition $\phi = \frac{\pi}{2} - \theta$ to point $B(\sec \phi, 2 \tan \phi)$, the corresponding normal equation is $x + y \csc(\frac{\pi}{2} - \theta) = 3 \sec(\frac{\pi}{2} - \theta)$, which results in $x + y \sec \theta = 3 \csc \theta$. Subtracting these two expressions isolates the y-coordinate of the intersection, as $y(\csc \theta - \sec \theta) = 3(\sec \theta - \csc \theta)$. Since $(\csc \theta - \sec \theta) = -(\sec \theta - \csc \theta)$, we determine that $y = -3$. With $\beta = -3$, the value of $(2\beta)^2$ is calculated as $(2 \times (-3))^2$, resulting in 36.

Identifying the parameters of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ reveals $a^{2}=25$ and $b^{2}=16$, which leads to a focal distance of $c = \sqrt{25-16}=3$. Because the hyperbola shares these foci at $(\pm 3, 0)$ and its axes coincide with those of the ellipse, we find the hyperbola's constant $a^{2}$ by substituting a focus point into the standard form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$, confirming $a^{2}=9$.

The eccentricity of the ellipse is $e_{e} = \sqrt{1-\frac{16}{25}} = \frac{3}{5}$. Given that the product of the eccentricities is one, the eccentricity of the hyperbola must be the reciprocal, $e_{h} = \frac{5}{3}$. Substituting this into the hyperbola's eccentricity formula $e_{h}^{2} = 1 + \frac{b^{2}}{a^{2}}$ results in $\frac{25}{9} = 1 + \frac{b^{2}}{9}$, which simplifies to $\frac{b^{2}}{9} = \frac{16}{9}$ and yields $b^{2}=16$. Consequently, the equation is $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.

The hyperbola defined by $x^2 - y^2 \sec^2 \theta = 10$ can be written in standard form as $\frac{x^2}{10} - \frac{y^2}{10 \cos^2 \theta} = 1$, which corresponds to an eccentricity of $e_H = \sqrt{1 + \frac{10 \cos^2 \theta}{10}} = \sqrt{1 + \cos^2 \theta}$. Similarly, the ellipse $x^2 \sec^2 \theta + y^2 = 5$ is expressed as $\frac{x^2}{5 \cos^2 \theta} + \frac{y^2}{5} = 1$. Since the denominator for $y^2$ is larger than that of $x^2$, the ellipse has its major axis along the $y$-axis with an eccentricity of $e_E = \sqrt{1 - \frac{5 \cos^2 \theta}{5}} = \sqrt{1 - \cos^2 \theta} = \sin \theta$.

Using the given condition $e_H = \sqrt{5} e_E$, the equality $\sqrt{1 + \cos^2 \theta} = \sqrt{5} \sin \theta$ holds. Squaring both sides produces $1 + \cos^2 \theta = 5 \sin^2 \theta$, and substituting the identity $\sin^2 \theta = 1 - \cos^2 \theta$ yields $1 + \cos^2 \theta = 5(1 - \cos^2 \theta)$, which simplifies to $1 + \cos^2 \theta = 5 - 5 \cos^2 \theta$. Rearranging this equation results in $6 \cos^2 \theta = 4$, or $\cos^2 \theta = \frac{2}{3}$. The length of the latus rectum of this ellipse is calculated using the formula $\frac{2 \cdot (\text{semi-minor axis})^2}{\text{semi-major axis}} = \frac{2(5 \cos^2 \theta)}{\sqrt{5}}$. Substituting the derived value of $\cos^2 \theta$ into this expression leads to $\frac{10(2/3)}{\sqrt{5}} = \frac{20}{3 \sqrt{5}} = \frac{4 \sqrt{5}}{3}$.

The eccentricity of a conic section is determined by the ratio of its axes, calculated as $\sqrt{1 - \frac{b^2}{a^2}}$ for an ellipse and $\sqrt{1 + \frac{b^2}{a^2}}$ for a hyperbola. For the ellipse $\frac{x^2}{18}+\frac{y^2}{4}=1$, the square of the eccentricity is $e_1^2 = 1-\frac{4}{18} = \frac{7}{9}$. For the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$, the square of the eccentricity is $e_2^2 = 1+\frac{4}{9} = \frac{13}{9}$.

Since the coordinate pair $(e_1, e_2)$ satisfies the curve defined by $15x^2 + 3y^2 = k$, substituting these eccentricity values into the expression yields:
$15\left(\frac{7}{9}\right) + 3\left(\frac{13}{9}\right) = k$
Calculating the terms results in $\frac{105}{9} + \frac{39}{9} = \frac{144}{9}$, which simplifies to $k = 16$.

The slope of a tangent line to a curve can be determined directly from the derivative or by utilizing the standard equation for a tangent at a specific point. Because the tangent line must be parallel to $2x - y = 0$, its slope must equal 2. For the hyperbola defined by $\frac{x^2}{4} - \frac{y^2}{2} = 1$, the tangent equation at a point $(x_1, y_1)$ is $\frac{x x_1}{4} - \frac{y y_1}{2} = 1$. Rearranging this into the slope-intercept form reveals that the slope is $\frac{x_1}{2y_1}$, which must be equal to 2, yielding the relationship $x_1 = 4y_1$.

Since the point $(x_1, y_1)$ must satisfy the hyperbola's equation, substituting $x_1 = 4y_1$ into $\frac{x_1^2}{4} - \frac{y_1^2}{2} = 1$ leads to:
$\frac{(4y_1)^2}{4} - \frac{y_1^2}{2} = 1$
This simplifies to $4y_1^2 - \frac{1}{2}y_1^2 = 1$, or $\frac{7}{2}y_1^2 = 1$, which establishes that $y_1^2 = \frac{2}{7}$. With this value, we evaluate the expression $x_1^2 + 5y_1^2$ by replacing $x_1^2$ with $(4y_1)^2$, resulting in $16y_1^2 + 5y_1^2$, or $21y_1^2$. Multiplying 21 by $\frac{2}{7}$ yields the final result of 6.

The eccentricity of the ellipse $\frac{x^2}{25} + \frac{y^2}{b^2} = 1$ is related to its parameters by $b^2 = 25(1 - e_1^2)$, whereas the eccentricity of the hyperbola $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$ is related by $b^2 = 16(e_2^2 - 1)$. Because both curves share the same value for $b^2$, we can equate these two expressions to establish the relationship $25(1 - e_1^2) = 16(e_2^2 - 1)$.

Given that the product of the eccentricities is unity, we substitute $e_2 = \frac{1}{e_1}$ into our equation, leading to $25(1 - e_1^2) = 16\left(\frac{1}{e_1^2} - 1\right)$, which simplifies to $25(1 - e_1^2) = 16\left(\frac{1 - e_1^2}{e_1^2}\right)$. Assuming $e_1$ is not unity, we divide both sides by $(1 - e_1^2)$ to find $25 = \frac{16}{e_1^2}$, which yields $e_1^2 = \frac{16}{25}$, resulting in $e_1 = \frac{4}{5}$ and consequently $e_2 = \frac{5}{4}$.

The distance between the foci is calculated using $2ae$ for both conics. For the ellipse, using $a = 5$ and $e_1 = \frac{4}{5}$, the distance is $\alpha = 2 \times 5 \times \frac{4}{5} = 8$. For the hyperbola, using $a = 4$ and $e_2 = \frac{5}{4}$, the distance is $\beta = 2 \times 4 \times \frac{5}{4} = 10$. Combining these results, the ordered pair $(\alpha, \beta)$ is $(8, 10)$.

The ellipse $3x^2 + 4y^2 = 12$ can be rewritten in standard form as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. With $a^2 = 4$ and $b^2 = 3$, the eccentricity is determined by $e = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$, making the foci, located at $(\pm ae, 0)$, equivalent to $(\pm 1, 0)$.

A hyperbola with a transverse axis length of $2a = \sqrt{2}$ possesses a semi-transverse axis $a = \frac{1}{\sqrt{2}}$, implying $a^2 = \frac{1}{2}$. Sharing these same foci with the ellipse means the distance from the center to each focus, $\sqrt{a^2 + b^2}$, must equal $1$. Squaring both sides gives $a^2 + b^2 = 1$, and substituting the known value $a^2 = \frac{1}{2}$ results in $b^2 = \frac{1}{2}$. The equation of the hyperbola is thus $\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$, which simplifies to $x^2 - y^2 = \frac{1}{2}$.

Substitution of the given coordinate pairs into this equation identifies which point does not reside on the curve. Point A satisfies the equation since $1^2 - (-\frac{1}{\sqrt{2}})^2 = 1 - \frac{1}{2} = \frac{1}{2}$, and point C follows similarly with $(\frac{1}{\sqrt{2}})^2 - 0 = \frac{1}{2}$. Testing point D gives $(-\sqrt{\frac{3}{2}})^2 - 1^2 = \frac{3}{2} - 1 = \frac{1}{2}$, which confirms it lies on the curve. In contrast, evaluating point B yields $(\sqrt{\frac{3}{2}})^2 - (\frac{1}{\sqrt{2}})^2 = \frac{3}{2} - \frac{1}{2} = 1$, which does not equal the required value of $\frac{1}{2}$.

Substituting the point $(3,3)$ directly into the hyperbola equation $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ provides the initial relationship $\frac{9}{a^{2}} - \frac{9}{b^{2}} = 1$. To define the geometry of the normal at $(3,3)$, we determine the tangent slope to be $\frac{b^{2}}{a^{2}}$, which implies the normal line has a slope of $-\frac{a^{2}}{b^{2}}$. Writing the equation of this normal as $y - 3 = -\frac{a^{2}}{b^{2}}(x - 3)$ and applying the condition that it intersects the x-axis at $(9,0)$ gives $-3 = -\frac{a^{2}}{b^{2}}(6)$, which simplifies to the ratio $b^{2} = 2a^{2}$. With this proportionality established, the eccentricity squared is found via $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + 2 = 3$. Substituting $b^{2} = 2a^{2}$ back into the first equation produces $\frac{9}{a^{2}} - \frac{9}{2a^{2}} = 1$, which reveals that $a^{2} = \frac{9}{2}$.