Ellipse – JEE Main MathematicsPractice Questions & PYQs

The eccentricity and the geometry of an ellipse are fundamentally connected by the relationship between its semi-axes, where for an ellipse with $a < b$ and major axis along the y-axis, the squared eccentricity is defined as $e^2 = 1 - \frac{a^2}{b^2}$. Substituting the given eccentricity $e = \frac{\sqrt{5}}{3}$ yields $e^2 = \frac{5}{9}$, leading to the relationship $1 - \frac{a^2}{b^2} = \frac{5}{9}$, which simplifies to $\frac{a^2}{b^2} = \frac{4}{9}$, or $b^2 = \frac{9}{4}a^2$.

Since the point (4, 3) must satisfy the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, substituting these coordinates results in $\frac{16}{a^2} + \frac{9}{b^2} = 1$. Replacing $b^2$ with $\frac{9}{4}a^2$ transforms this into $\frac{16}{a^2} + \frac{9}{(\frac{9}{4}a^2)} = 1$, which further reduces to $\frac{16}{a^2} + \frac{4}{a^2} = 1$. Solving this equation gives $a^2 = 20$, and subsequently $b^2 = \frac{9}{4}(20) = 45$, so the semi-major axis length is $b = \sqrt{45} = 3\sqrt{5}$. The length of the latus rectum for this ellipse is calculated using the formula $\frac{2a^2}{b}$, which becomes $\frac{2(20)}{3\sqrt{5}} = \frac{40}{3\sqrt{5}}$, resulting in $\frac{8\sqrt{5}}{3}$ after rationalizing the denominator.

The locus of a point $B(x, y)$ such that a circle with diameter $AB$ touches the circle $x^{2}+y^{2}=36$ internally relies on the distance relationship between the two centers. With $A$ located at $(3, 0)$, the variable circle is centered at $(\frac{x+3}{2}, \frac{y}{2})$ and has a radius $r = \frac{1}{2}\sqrt{(x-3)^2 + y^2}$. Internal tangency implies that the distance between the origin and the variable center equals the difference between the fixed circle radius $6$ and the radius $r$.

This relationship is written as $\sqrt{(\frac{x+3}{2})^2 + (\frac{y}{2})^2} = 6 - \frac{1}{2}\sqrt{(x-3)^2 + y^2}$. Multiplying by two simplifies this to $\sqrt{(x+3)^2 + y^2} = 12 - \sqrt{(x-3)^2 + y^2}$, which rearranges into $\sqrt{(x+3)^2 + y^2} + \sqrt{(x-3)^2 + y^2} = 12$. This form defines an ellipse with foci at $(-3, 0)$ and $(3, 0)$ where the constant sum of focal distances $2a = 12$ gives $a = 6$. Because the distance between the foci is $2c = 6$, the value of $c$ is $3$, yielding an eccentricity $e = \frac{3}{6} = \frac{1}{2}$, which results in $72e^{2} = 72 \times \frac{1}{4} = 18$.

Because the center, focus, and vertex all share a fixed y-coordinate of $-2$, the major axis of this ellipse must lie along a horizontal line. We determine the semi-major axis $a$ by calculating the distance from the center $(1, -2)$ to the vertex $(5, -2)$, which gives $a = 5 - 1 = 4$, and find the focal distance $c$ by measuring the distance from the center to the focus $(3, -2)$, which yields $c = 3 - 1 = 2$. With these values established, the square of the semi-minor axis $b^2$ is found through the identity $b^2 = a^2 - c^2$, resulting in $b^2 = 4^2 - 2^2 = 12$. Substituting these variables into the standard formula for the length of the latus rectum, $\frac{2b^2}{a}$, provides a final value of $\frac{2(12)}{4} = 6$.

The geometry of an ellipse is defined by its semi-major axis $a$ and eccentricity $e$, which allow us to locate the focus $S$ at the coordinates $(ae, 0)$. Since the focus $S$ is at $(4,0)$ and the eccentricity is $\frac{4}{5}$, we find $ae = 4$, which leads to $a = 5$. Utilizing the standard relationship $b^2 = a^2(1 - e^2)$, we calculate the semi-minor axis squared as $b^2 = 25(1 - \frac{16}{25}) = 9$, providing the ellipse equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Substituting the coordinates of point $P(3, \alpha)$ into this equation gives $\frac{3^2}{25} + \frac{\alpha^2}{9} = 1$, which simplifies to $\frac{9}{25} + \frac{\alpha^2}{9} = 1$. Rearranging this expression reveals $\frac{\alpha^2}{9} = \frac{16}{25}$, which means $\alpha^2 = \frac{144}{25}$, or $\alpha = \pm \frac{12}{5}$. The triangle $\triangle POS$ has vertices at the origin $O(0,0)$, the focus $S(4,0)$, and the point $P(3, \pm \frac{12}{5})$. Treating the segment $OS$ on the x-axis as the base with length 4, the corresponding height of the triangle is the absolute value of the y-coordinate of $P$, which is $\frac{12}{5}$. Calculating the area as $\frac{1}{2} \times \text{base} \times \text{height}$ yields $\frac{1}{2} \times 4 \times \frac{12}{5}$, resulting in $\frac{24}{5}$.

The coordinates of a triangle's centroid are defined as the arithmetic mean of the coordinates of its three vertices, $P, Q,$ and $R$. We define $P = (3 \cos \alpha, 2 \sin \alpha)$, represent $Q$ in its parametric form as $(7 + 4 \cos \theta, 7 + 4 \sin \theta)$ based on the circle equation $(x - 7)^2 + (y - 7)^2 = 16$, and denote $R$ as $(x_R, 5 - x_R)$ since it resides on the line $x + y = 5$. Equating the centroid's vertical component to the given value gives:

$\frac{2 \sin \alpha + 7 + 4 \sin \theta + 5 - x_R}{3} = 3 + \frac{2}{3} \sin \alpha$

Simplifying this equation results in $x_R = 3 + 4 \sin \theta$. Performing a similar calculation for the horizontal component,

$\frac{3 \cos \alpha + 7 + 4 \cos \theta + x_R}{3} = 2 + \cos \alpha$

leads to $x_R = -1 - 4 \cos \theta$. Setting these two expressions for $x_R$ equal to each other gives $3 + 4 \sin \theta = -1 - 4 \cos \theta$, which simplifies further to the condition $\sin \theta + \cos \theta = -1$. Squaring this identity yields $1 + 2 \sin \theta \cos \theta = 1$, or $\sin \theta \cos \theta = 0$, implying that either $\sin \theta = 0$ or $\sin \theta = -1$. Substituting these values into $y_R = 5 - x_R = 5 - (3 + 4 \sin \theta) = 2 - 4 \sin \theta$ provides potential $y$-coordinates of $2$ and $6$, resulting in a sum of $8$.

An ellipse follows the structure $\frac{x^{2}}{B^{2}} + \frac{y^{2}}{A^{2}} = 1$ with its major axis along the y-axis when the denominator of the $y^{2}$ term is strictly greater than the denominator of the $x^{2}$ term. Since $f$ is a strictly decreasing function, the relationship $f(u) > f(v)$ implies that the input $u$ must be strictly less than the input $v$. Applying this property to the equation provided, the condition that the major axis is vertical implies $f(3a+15) > f(a^{2}+7a+3)$, which simplifies to the inequality $3a+15 < a^{2}+7a+3$.

Moving all terms to one side of the inequality yields $a^{2}+4a-12 > 0$, which factors into the expression $(a+6)(a-2) > 0$. This quadratic inequality is satisfied for values of $a$ outside the interval between the roots $-6$ and $2$, meaning $a \in (-\infty , -6) \cup (2, \infty )$. Because the set of all possible values is defined as $R-[\alpha,\beta]$, the excluded interval is $[-6, 2]$, identifying $\alpha$ as $-6$ and $\beta$ as $2$. Consequently, calculating $\alpha^{2}+\beta^{2}$ results in $(-6)^{2}+(2)^{2}$, which sums to $36+4$, yielding $40$.

For an ellipse centred at the origin, the directrix is defined by the equation $x = a/e$. Given the directrix $x = 9$ and the eccentricity $e = 1/3$, we determine $a = 3$. The semi-minor axis squared is calculated using $b^2 = a^2(1 - e^2)$, which results in $b^2 = 9(1 - 1/9) = 8$. Consequently, the equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{8} = 1$. Because the focus is located at $P(ae, 0)$, we identify the focus at $P(1, 0)$.

The equation of a chord with a specific midpoint $(h, k)$ is represented by the formula $T = S_1$, where $T = \frac{xh}{9} + \frac{yk}{8} - 1$ and $S_1 = \frac{h^2}{9} + \frac{k^2}{8} - 1$. Since the chord passes through the focus $P(1, 0)$, we substitute $x = 1$ and $y = 0$ into this condition to establish the relationship between the coordinates of the midpoint. This substitution yields $\frac{h}{9} - 1 = \frac{h^2}{9} + \frac{k^2}{8} - 1$, which simplifies to $\frac{h}{9} = \frac{h^2}{9} + \frac{k^2}{8}$. Multiplying by 72 gives $8h = 8h^2 + 9k^2$, or $9k^2 = 8h - 8h^2$. Writing this in terms of general coordinates $x$ and $y$ gives the final locus $9y^2 = 8x(1 - x)$.

The fundamental property of an ellipse states that the sum of the distances from any point on the curve to the two foci is constant and equal to the length of the major axis, which in this case is $2a = 10$. The given algebraic expression $(SP)^2+(S'P)^2-SP\cdot S'P=37$ can be simplified by recognizing that $SP^2+S'P^2$ is equivalent to $(SP+S'P)^2-2SP\cdot S'P$. Substituting the known sum of the focal distances into this identity leads to $(10)^2 - 3SP\cdot S'P = 37$, which reveals that the product of the distances $SP\cdot S'P$ must be $21$.

For the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$, the focal distances from a point $(\alpha, \beta)$ are $a+e\alpha$ and $a-e\alpha$, where $a=5$ and the eccentricity $e$ is $\sqrt{1-\frac{9}{25}} = \frac{4}{5}$. Multiplying these expressions gives $a^2 - e^2\alpha^2$, or $25 - \frac{16}{25}\alpha^2$, which we equate to $21$ to find $\alpha^2 = \frac{25}{4}$. Using the ellipse equation $\frac{\alpha^2}{25}+\frac{\beta^2}{9}=1$, we substitute $\alpha^2$ to determine $\frac{\beta^2}{9} = 1 - \frac{1}{4} = \frac{3}{4}$, resulting in $\beta^2 = \frac{27}{4}$. Combining these values yields $\alpha^2+\beta^2 = \frac{25}{4}+\frac{27}{4}$, which simplifies to $13$.

Curves passing through the intersection points of two ellipses are represented by a linear combination of the two equations, expressed as $S_1 + \lambda S_2 = 0$. For this combined curve to be a circle, the coefficients of the quadratic $x^2$ and $y^2$ terms must be equal, as this is the defining requirement for the equation of a circle.

Combining the given ellipses $x^2 + 2y^2 - 6x - 12y + 23 = 0$ and $4x^2 + 2y^2 - 20x - 12y + 35 = 0$ with an arbitrary parameter $\lambda$ results in the equation $(1 + 4\lambda)x^2 + (2 + 2\lambda)y^2 - (6 + 20\lambda)x - (12 + 12\lambda)y + (23 + 35\lambda) = 0$. Equating the coefficients of $x^2$ and $y^2$ leads to $1 + 4\lambda = 2 + 2\lambda$, which yields $\lambda = 1/2$. Replacing $\lambda$ with this value in the combined equation produces $3x^2 + 3y^2 - 16x - 18y + 81/2 = 0$, and normalizing this by dividing through by $3$ gives $x^2 + y^2 - \frac{16}{3}x - 6y + \frac{27}{2} = 0$.

Identifying the center of this circle requires comparing the general form $x^2 + y^2 + 2gx + 2fy + c = 0$ with our equation, revealing $g = -8/3$ and $f = -3$, which places the center $(a, b)$ at $(8/3, 3)$. The squared radius $r^2$ is determined by evaluating $g^2 + f^2 - c$, calculated as $(-8/3)^2 + (-3)^2 - 27/2 = 64/9 + 9 - 27/2$. Finding a common denominator for this expression results in $r^2 = 47/18$. Determining the value of $ab + 18r^2$ involves multiplying $a$ and $b$ to get $8$, then adding $18 \times (47/18)$, which results in the calculation $8 + 47 = 55$.

For an ellipse with eccentricity $e$ and semi-major axis $a$, the length of the latus rectum is $2a(1-e^2)$ and the distance between the foci is $2ae$. Given $e = \frac{4}{5}$ and the distance between the foci of $E_1$ is $2ae = 8$, we determine $a = 5$, leading to a latus rectum length $l_1 = 2(5)(1 - \frac{16}{25}) = 10(\frac{9}{25}) = \frac{18}{5}$. Using the given relation $2l_1^2 = 9l_2$, we find $2(\frac{18}{5})^2 = 9l_2$, which simplifies to $l_2 = \frac{2 \times 324}{25 \times 9} = \frac{72}{25}$. Since $E_2$ has a vertical major axis $B$ and eccentricity $e = \frac{4}{5}$, its latus rectum length is $l_2 = 2B(1-e^2) = 2B(1 - \frac{16}{25}) = \frac{18B}{25}$. Equating the two expressions for $l_2$, we have $\frac{18B}{25} = \frac{72}{25}$, which gives $B = 4$. Consequently, the distance between the foci of $E_2$ is $2Be = 2(4)(\frac{4}{5}) = \frac{32}{5}$.

The eccentricity $e$ is determined by finding the vertex of the quadratic function $f(t) = -t^2 + 2t - \frac{3}{4}$, which occurs at $t = 1$ and yields a maximum value of $e = f(1) = \frac{1}{4}$. Given the latus rectum length formula:
$\frac{2b^2}{a} = 30$
we establish the relation $b^2 = 15a$. Substituting this into the eccentricity identity $e^2 = 1 - \frac{b^2}{a^2}$ with $e = \frac{1}{4}$ results in the equation $\frac{1}{16} = 1 - \frac{15a}{a^2}$, which simplifies to $\frac{15}{a} = \frac{15}{16}$ and yields $a = 16$. Finally, with $a = 16$, the values $b^2 = 240$ and $a^2 = 256$ lead to $a^2 + b^2 = 496$.

Since $(h, k)$ lies on the circle $x^2+y^2=4$, we relate these coordinates to the ellipse coordinates $(x, y) = (2h+1, 3k+2)$ by substituting $h = \frac{x-1}{2}$ and $k = \frac{y-2}{3}$ into the circle equation to obtain $\left(\frac{x-1}{2}\right)^2 + \left(\frac{y-2}{3}\right)^2 = 4$. Expanding this expression and dividing by $4$ transforms the equation into the standard ellipse form $\frac{(x-1)^2}{16} + \frac{(y-2)^2}{36} = 1$. Identifying the squared semi-axes as $a^2 = 16$ and $b^2 = 36$, the eccentricity squared is determined by $e^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{16}{36} = \frac{20}{36} = \frac{5}{9}$. Evaluating the required value $\frac{5}{e^2}$ thus yields $5 \div \frac{5}{9}$, which results in $9$.

The tangency condition for a line with slope $m$ and y-intercept $c$ relative to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$. Converting the ellipse $3x^2 + 4y^2 = 1$ to standard form $\frac{x^2}{1/3} + \frac{y^2}{1/4} = 1$ identifies the squared semi-axes as $a^2 = 1/3$ and $b^2 = 1/4$. Rearranging the line $\alpha x + 4y = \sqrt{7}$ into $y = -\frac{\alpha}{4}x + \frac{\sqrt{7}}{4}$ reveals the slope $m = -\frac{\alpha}{4}$ and the intercept $c = \frac{\sqrt{7}}{4}$.

Substituting these values into the tangency equation yields $\frac{7}{16} = \frac{1}{3} \cdot \frac{\alpha^2}{16} + \frac{1}{4}$. Simplifying this expression leads to $\frac{\alpha^2}{48} = \frac{3}{16}$, which results in $\alpha^2 = 9$. Selecting $\alpha = 3$ ensures the point of contact $P$ lies in the first quadrant, establishing the tangent line as $3x + 4y = \sqrt{7}$.

The coordinates of the contact point $(x_1, y_1)$ must satisfy the tangent equation $3x x_1 + 4y y_1 = 1$ when compared against the form $\frac{3}{\sqrt{7}}x + \frac{4}{\sqrt{7}}y = 1$. This comparison shows $3x_1 = \frac{3}{\sqrt{7}}$ and $4y_1 = \frac{4}{\sqrt{7}}$, which simplifies to $x_1 = \frac{1}{\sqrt{7}}$ and $y_1 = \frac{1}{\sqrt{7}}$. The focal distances from point $P$ are given by $a \pm ex_1$. With eccentricity $e = \sqrt{1 - \frac{1/4}{1/3}} = \frac{1}{2}$ and $a = \frac{1}{\sqrt{3}}$, the calculation yields $\frac{1}{\sqrt{3}} \pm \frac{1}{2} \cdot \frac{1}{\sqrt{7}}$, resulting in focal distances of $\frac{1}{\sqrt{3}} \pm \frac{1}{2\sqrt{7}}$.

Finding the angle subtended by a chord at the center of an ellipse is equivalent to determining the difference in the polar angles of the chord's endpoints relative to the origin. To find the points of intersection for the line $y = x + 1$ and the ellipse $\frac{x^2}{2} + y^2 = 1$, substitute the line equation into the ellipse equation to obtain $\frac{x^2}{2} + (x + 1)^2 = 1$. Multiplying by 2 and simplifying yields $x^2 + 2(x^2 + 2x + 1) = 2$, which reduces to $3x^2 + 4x = 0$. Solving this quadratic provides the x-coordinates $x = 0$ and $x = -\frac{4}{3}$, which correspond to the intersection points $A(0, 1)$ and $B\left(-\frac{4}{3}, -\frac{1}{3}\right)$.

Because the ellipse is centered at the origin $O(0, 0)$, the vector $\overrightarrow{OA}$ lies along the positive y-axis, making an angle of $\frac{\pi}{2}$ with the positive x-axis. The point $B$ resides in the third quadrant where both coordinates are negative, meaning the slope of the vector $\overrightarrow{OB}$ is $\frac{-1/3}{-4/3} = \frac{1}{4}$. The polar angle for this vector in the third quadrant is $\pi + \tan^{-1}\left(\frac{1}{4}\right)$. Subtracting the angle of $\overrightarrow{OA}$ from the angle of $\overrightarrow{OB}$ gives the central angle $\left(\pi + \tan^{-1}\left(\frac{1}{4}\right)\right) - \frac{\pi}{2}$, which simplifies to $\frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$.

The length of the latus rectum of an ellipse is given by $\frac{2b^2}{a}$, where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis. Given the foci $(2, 5)$ and $(2, -3)$, the distance between them is $2ae = \sqrt{(2-2)^2 + (5 - (-3))^2} = 8$, which simplifies to $ae = 4$. With the eccentricity $e = \frac{4}{5}$, the semi-major axis $a$ is calculated as $a = \frac{4}{4/5} = 5$. The square of the semi-minor axis is determined by the relation $b^2 = a^2(1 - e^2) = 25(1 - \frac{16}{25}) = 25(\frac{9}{25}) = 9$. Finally, substituting these values into the expression for the latus rectum results in $\frac{2(9)}{5} = \frac{18}{5}$.

Any chord of an ellipse with a known midpoint $(x_1, y_1)$ is determined by the equation $T = S_1$, where $T$ represents the tangent-style form $\frac{x x_1}{a^2} + \frac{y y_1}{b^2} - 1$ and $S_1$ is the value obtained by substituting the midpoint directly into the ellipse equation.

For the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ with midpoint $(\sqrt{2}, \frac{4}{3})$, the chord equation is established as:
$\frac{x\sqrt{2}}{9} + \frac{y(4/3)}{4} - 1 = \frac{(\sqrt{2})^2}{9} + \frac{(4/3)^2}{4} - 1$
Simplifying this expression yields $\frac{x\sqrt{2}}{9} + \frac{y}{3} = \frac{2}{3}$, which rearranges to $\sqrt{2}x + 3y = 6$. Substituting $x = \frac{6 - 3y}{\sqrt{2}}$ into the original ellipse equation results in the quadratic $3y^2 - 8y + 4 = 0$, giving intersection points at $y = 2$ and $y = 2/3$. These correspond to the coordinates $(0, 2)$ and $(2\sqrt{2}, 2/3)$, and calculating the distance between them gives:
$\sqrt{(2\sqrt{2} - 0)^2 + \left(\frac{2}{3} - 2\right)^2} = \sqrt{8 + \frac{16}{9}} = \frac{2\sqrt{22}}{3}$
Comparing this result to $\frac{2\sqrt{\alpha}}{3}$ leads directly to $\alpha = 22$.

For any point on an ellipse with semi-major axis $a$, the sum of the distances to the foci $SP$ and $S'P$ remains constant at $2a$. In the given equation $\frac{x^2}{18} + \frac{y^2}{9} = 1$, we identify $a = \sqrt{18} = 3\sqrt{2}$, which fixes the sum $SP + S'P$ at $6\sqrt{2}$.

The product $SP \cdot S'P$ is maximized when the point $P$ is at the ends of the minor axis, where $SP = S'P = a$, resulting in a maximum value of $(3\sqrt{2})^2 = 18$. The minimum product occurs at the vertices of the major axis, where the distances are $a - ae$ and $a + ae$; given $e = \frac{1}{\sqrt{2}}$, the focal distance $ae$ is $3$. Calculating the product at these vertices yields $(3\sqrt{2} - 3)(3\sqrt{2} + 3) = 18 - 9 = 9$. Adding the maximum and minimum values together gives $18 + 9 = 27$.

The circle equation $x^2 + y^2 - 2x - 4y - 11 = 0$ can be rewritten as $(x-1)^2 + (y-2)^2 = 16$, which identifies the center at (1, 2) and a radius of 4. Inserting the coordinates of this center into the ellipse equation $3x^2 + py^2 = 4$ yields $3(1)^2 + p(2)^2 = 4$, leading to $4p = 1$ and therefore $p = 1/4$. This defines the ellipse as $3x^2 + \frac{y^2}{4} = 4$, or in standard form, $\frac{x^2}{4/3} + \frac{y^2}{16} = 1$. Since the denominator under $y^2$ is larger, the major axis lies along the y-axis, and the eccentricity is determined by $e = \sqrt{1 - \frac{4/3}{16}} = \sqrt{\frac{11}{12}}$.

The foci are at $(0, \pm ae)$, which evaluates to $(0, \pm 2 \sqrt{\frac{11}{3}})$. Given the focal distances $f_1$ and $f_2$ from the center point (1, 2), their product is found by calculating $f_1 f_2 = \sqrt{(1^2 + (2 - 2\sqrt{\frac{11}{3}})^2)(1^2 + (2 + 2\sqrt{\frac{11}{3}})^2)}$. Expanding this product simplifies to $\sqrt{(\frac{59}{3})^2 - 64(\frac{11}{3})}$, which results in $\sqrt{\frac{3481}{9} - \frac{2112}{9}} = 37/3$. Substituting this into the final expression $6 f_1 f_2 - r$, the calculation $6(37/3) - 4$ yields $74 - 4$, resulting in 70.

The circle passing through the foci $F_1(-c, 0)$ and $F_2(c, 0)$ centered at the origin has the radius $R = c$, defined by $x^2 + y^2 = c^2$. Because the intersection point $P(x_0, y_0)$ satisfies both $x^2 + y^2 = c^2$ and the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we substitute $y_0^2 = c^2 - x_0^2$ to obtain $\frac{x_0^2}{a^2} + \frac{c^2 - x_0^2}{b^2} = 1$. Rearranging this leads to $x_0^2(\frac{1}{a^2} - \frac{1}{b^2}) = \frac{b^2 - c^2}{b^2}$, which simplifies to $x_0^2(\frac{-c^2}{a^2 b^2}) = \frac{a^2 - 2c^2}{b^2}$ given $b^2 = a^2 - c^2$, yielding $x_0^2 = \frac{a^2(2c^2 - a^2)}{c^2}$.