Ellipse – JEE Main MathematicsPractice Questions & PYQs

The geometry of this problem relies on determining the intersection of a linear constraint with a circular boundary derived from the properties of the given ellipse. The ellipse $x^2 + 9y^2 = 9$ can be expressed as $\frac{x^2}{3^2} + \frac{y^2}{1^2} = 1$, identifying its major axis as the segment between $x=-3$ and $x=3$. Since this major axis forms the diameter of the circle $C$, the circle is centered at the origin $O(0, 0)$ with a radius of $3$, resulting in the equation $x^2 + y^2 = 9$.

To find points $A$ and $B$, we set the coordinates to identify the axis intercepts. Setting $y=0$ in the ellipse equation gives $x^2=9$, yielding $A(3, 0)$ on the positive $x$-axis. Setting $x=0$ results in $9y^2=9$, yielding $y=1$ and thus $B(0, 1)$ on the positive $y$-axis. The line passing through these points is represented by the intercept form $\frac{x}{3} + \frac{y}{1} = 1$, which simplifies to $x + 3y = 3$.

Point $P$ is the intersection of the line $x = 3 - 3y$ and the circle $x^2 + y^2 = 9$. Substituting the expression for $x$ into the circle equation yields $(3 - 3y)^2 + y^2 = 9$. Expanding this expression leads to $9(1 - 2y + y^2) + y^2 = 9$, which simplifies to $10y^2 - 18y = 0$. Solving this quadratic equation gives $y=0$ or $y=\frac{9}{5}$, where $y=0$ corresponds to point $B$. Using $y=\frac{9}{5}$, the corresponding $x$-coordinate is $x = 3 - 3(\frac{9}{5}) = -\frac{12}{5}$, identifying $P$ as $(-\frac{12}{5}, \frac{9}{5})$.

The area of the triangle with vertices $A(3, 0)$, $O(0, 0)$, and $P(-\frac{12}{5}, \frac{9}{5})$ is calculated by using the segment $OA$ as the base along the $x$-axis, which has a length of $3$. The altitude of the triangle is the absolute value of the $y$-coordinate of $P$, which is $\frac{9}{5}$. Applying the area formula $\frac{1}{2} \times \text{base} \times \text{height}$ yields $\frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10}$. Given this area is equal to $\frac{m}{n}$ with $m=27$ and $n=10$, the difference $m-n$ is $17$.

By applying the principle of inclusion and exclusion, we find the number of people who speak both languages by adding the total counts of English and Hindi speakers and subtracting the total number of people, which yields $75 + 40 - 100 = 15$ persons. Consequently, the number of persons speaking only English is $\alpha = 75 - 15 = 60$, and the number of persons speaking only Hindi is $\beta = 40 - 15 = 25$.

The given ellipse equation can be expressed in standard form through division and substitution:
$25 (\beta^2 x^2 + \alpha^2 y^2) = \alpha^2 \beta^2$
Dividing both sides by the product $\alpha^2 \beta^2$ leads to the equation:
$\frac{25 x^2}{\alpha^2} + \frac{25 y^2}{\beta^2} = 1$
Substituting the calculated values of $\alpha = 60$ and $\beta = 25$ yields $\frac{25 x^2}{60^2} + \frac{25 y^2}{25^2} = 1$, which simplifies to $\frac{x^2}{(60/5)^2} + \frac{y^2}{(25/5)^2} = 1$, or $\frac{x^2}{12^2} + \frac{y^2}{5^2} = 1$. With the semi-major axis $a = 12$ and the semi-minor axis $b = 5$, the eccentricity is determined by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$. Evaluating this expression results in $e = \sqrt{1 - \frac{25}{144}} = \sqrt{\frac{119}{144}}$, which simplifies to $\frac{\sqrt{119}}{12}$.

The equation of the normal to the ellipse $\frac{x^2}{4} + \frac{y^2}{b^2} = 1$ at a point with eccentric angle $\theta$ is $2x \sec \theta - by \csc \theta = 4 - b^2$. The perpendicular distance from the origin $(0,0)$ to this line is expressed as $d = \frac{|4 - b^2|}{\sqrt{4 \sec^2 \theta + b^2 \csc^2 \theta}}$. Given $b < 2$, the term $4 - b^2$ is positive, so the maximum value of the distance corresponds to the minimum value of the denominator $f(\theta) = 4 \sec^2 \theta + b^2 \csc^2 \theta$. This expression reaches its minimum when $\tan^2 \theta = \frac{b}{2}$, which, upon substitution back into the denominator, results in $\sqrt{4(1 + b/2) + b^2(1 + 2/b)} = b + 2$. The maximum distance is therefore $\frac{4 - b^2}{b + 2} = 2 - b$. Equating this maximum distance to 1 yields $2 - b = 1$, so $b = 1$. With the semi-major axis $a = 2$ and semi-minor axis $b = 1$, the eccentricity is calculated as $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.

The largest circle centered at $(2,0)$ inscribed in the ellipse $\frac{x^2}{36} + \frac{y^2}{9} = 1$ is tangent to the ellipse at the point where the normal passes through the circle's center. For any point $(6 \cos \theta, 3 \sin \theta)$ on the ellipse, the normal equation is $6x \sec \theta - 3y \csc \theta = 27$. Substituting the center coordinates $(2,0)$ into this equation reveals $12 \sec \theta = 27$, which yields $\sec \theta = \frac{9}{4}$, $\cos \theta = \frac{4}{9}$, and $\sin \theta = \frac{\sqrt{65}}{9}$.

The point of tangency is therefore $P\left(\frac{8}{3}, \frac{\sqrt{65}}{3}\right)$. The squared radius $r^2$ is the squared distance between $(2,0)$ and $P$, calculated as $(\frac{8}{3} - 2)^2 + (\frac{\sqrt{65}}{3})^2$, simplifying to $\frac{4}{9} + \frac{65}{9} = \frac{69}{9}$. Multiplying this by $12$ gives $12 \times \frac{69}{9} = 92$.

An ellipse centered at $(1,0)$ with its major axis along the $x$-axis follows the standard equation $\frac{(x-1)^2}{a^2} + \frac{y^2}{b^2} = 1$. The length of the latus rectum is defined as $\frac{2b^2}{a}$, which is given as $\frac{1}{2}$, leading to the relationship $4b^2 = a$.

The minor axis endpoints are located at $(1, \pm b)$ and the foci are at $(1 \pm ae, 0)$. Since the minor axis subtends an angle of $60^{\circ}$ at the foci, the triangle formed by the center, a focus, and an endpoint of the minor axis contains an angle of $30^{\circ}$ at the focus. This implies $\tan(30^{\circ}) = \frac{b}{ae} = \frac{1}{\sqrt{3}}$, which squares to $\frac{b^2}{a^2 e^2} = \frac{1}{3}$. Substituting the identity $a^2 e^2 = a^2 - b^2$ results in $3b^2 = a^2 - b^2$, or $4b^2 = a^2$.

Equating the two derived expressions for $4b^2$ gives $a = a^2$, and since the length $a$ must be positive, $a = 1$. This leads to $4b^2 = 1$, so $b^2 = \frac{1}{4}$ and $b = \frac{1}{2}$. The major axis has a length of $2a = 2$, and the minor axis has a length of $2b = 1$. The square of the sum of these lengths is $(2 + 1)^2$, which equals $9$.

Normalizing the ellipse equation $15 x^2 + 19 y^2 = 285$ by dividing by 285 yields $\frac{x^2}{19} + \frac{y^2}{15} = 1$. A tangent line to this ellipse with slope $m$ is given by $y = mx \pm \sqrt{19m^2 + 15}$. Since the circle is concentric with the ellipse at the origin and has a radius of 4, the perpendicular distance from the origin to this tangent line must equal 4. Applying the standard distance formula for the line $mx - y \pm \sqrt{19m^2 + 15} = 0$, the relationship $\frac{\sqrt{19m^2 + 15}}{\sqrt{m^2 + 1}} = 4$ must hold. Squaring both sides and simplifying results in $19m^2 + 15 = 16(m^2 + 1)$, which reduces to $3m^2 = 1$ or $m^2 = \frac{1}{3}$. This result indicates the tangent line makes an angle of $\frac{\pi}{6}$ with the major axis of the ellipse. Because the minor axis is perpendicular to the major axis, the angle the tangent makes with the minor axis is $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.

The ellipse $9 x^2+16 y^2=144$ can be rewritten in standard form as $\frac{x^2}{16}+\frac{y^2}{9}=1$, identifying its semi-axes as $4$ and $3$. A tangent line drawn to this ellipse at an arbitrary point $(4 \cos \theta, 3 \sin \theta)$ follows the equation $\frac{x \cos \theta}{4}+\frac{y \sin \theta}{3}=1$. By setting $y=0$ to find the $x$-intercept $A(4 \sec \theta, 0)$ and setting $x=0$ to find the $y$-intercept $B(0, 3 \csc \theta)$, we establish the positions of the two endpoints.

The square of the distance between points $A$ and $B$ is given by $16 \sec^2 \theta + 9 \csc^2 \theta$. By applying the trigonometric identities $\sec^2 \theta = 1 + \tan^2 \theta$ and $\csc^2 \theta = 1 + \cot^2 \theta$, this expression becomes $16(1 + \tan^2 \theta) + 9(1 + \cot^2 \theta)$, which simplifies to $25 + 16 \tan^2 \theta + 9 \cot^2 \theta$. Invoking the Arithmetic Mean-Geometric Mean inequality, the minimum value of $16 \tan^2 \theta + 9 \cot^2 \theta$ is $2 \sqrt{16 \cdot 9}$, which equals $24$. Adding this value to the remaining constant $25$ results in a minimum squared length of $49$, and calculating the square root gives a minimum length of $7$ for the segment $AB$.

A family of lines passing through the intersection of the lines $b x+10 y-8=0$ and $2 x-3 y=0$ is represented by the equation $(b x+10 y-8)+\lambda(2 x-3 y)=0$, where $\lambda$ is a parameter. Because the line passes through the point $(1,1)$, we substitute these coordinates into the equation to find $\lambda$, leading to $(b+10-8)+\lambda(2-3)=0$, which simplifies to $b+2-\lambda=0$ or $\lambda=b+2$. Substituting this value back into the family equation gives $(b+2(b+2))x+(10-3(b+2))y-8=0$, which simplifies to $(3b+4)x+(4-3b)y-8=0$.

For this line to touch the circle $17(x^2+y^2)=16$, its perpendicular distance from the origin $(0,0)$ must equal the radius of the circle, $R=\sqrt{\frac{16}{17}}=\frac{4}{\sqrt{17}}$. Using the distance formula for the line $(3b+4)x+(4-3b)y-8=0$, we set $\frac{|-8|}{\sqrt{(3b+4)^2+(4-3b)^2}}=\frac{4}{\sqrt{17}}$. Squaring both sides of this equation results in $\frac{64}{(3b+4)^2+(4-3b)^2}=\frac{16}{17}$, which simplifies to $\frac{4}{18b^2+32}=\frac{1}{17}$. Solving this gives $68=18b^2+32$, which yields $18b^2=36$ and thus $b^2=2$. For an ellipse with the equation $\frac{x^2}{5}+\frac{y^2}{b^2}=1$, the eccentricity $e$ is determined by $e=\sqrt{1-\frac{b^2}{5}}$. Substituting $b^2=2$ gives $e=\sqrt{1-\frac{2}{5}}=\sqrt{\frac{3}{5}}$.

An ellipse defined by the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ possesses an eccentricity $e$, which relates the lengths of the semi-axes via $b^2 = a^2(1 - e^2)$. Since the ellipse passes through the coordinate $(-4\sqrt{\frac{2}{5}}, 3)$, substituting these values into the standard equation generates a relationship between the denominators $a^2$ and $b^2$. Plugging in $x = -4\sqrt{\frac{2}{5}}$ and $y = 3$ gives the following equation:

$\frac{32}{5a^2} + \frac{9}{b^2} = 1$

Given $e = \frac{1}{4}$, the eccentricity formula $b^2 = a^2(1 - \frac{1}{16})$ implies that $b^2 = \frac{15}{16}a^2$, which can be rearranged as $a^2 = \frac{16b^2}{15}$. Substituting this into the previous equation yields:

$\frac{32}{5(\frac{16b^2}{15})} + \frac{9}{b^2} = 1$

After simplifying the first term, we obtain $\frac{6}{b^2} + \frac{9}{b^2} = 1$, which leads to $\frac{15}{b^2} = 1$, meaning $b^2 = 15$. Consequently, $a^2 = 16$. The required sum $a^2 + b^2$ is 31.

The minimum distance between two curves is found along a common normal segment, and since $y=e^x$ and $y=\log_e x$ are symmetric reflections across $y=x$, the shortest path between them corresponds to the distance between points where the tangent slope equals 1. For $y=e^x$, the derivative $e^x=1$ identifies the point $(0,1)$, while for $y=\log_e x$, the derivative $\frac{1}{x}=1$ identifies the point $(1,0)$, making the distance $a = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{2}$, which means $a^2 = 2$.

The ellipse $\frac{x^2}{25a^2} + \frac{y^2}{25} = 1$ has an eccentricity $e_1 = \sqrt{1 - \frac{25}{25a^2}} = \sqrt{1 - \frac{1}{a^2}}$, and the hyperbola $\frac{x^2}{5} - \frac{y^2}{5/a^2} = 1$ has an eccentricity $e_2 = \sqrt{1 + \frac{5/a^2}{5}} = \sqrt{1 + \frac{1}{a^2}}$. The relationship $e_1 = b e_2$ implies $e_1^2 = b^2 e_2^2$, or $1 - \frac{1}{a^2} = b^2 \left(1 + \frac{1}{a^2}\right)$. Substituting $a^2 = 2$ into this equation gives $1 - \frac{1}{2} = b^2 \left(1 + \frac{1}{2}\right)$, which simplifies to $\frac{1}{2} = b^2 \left(\frac{3}{2}\right)$, resulting in $b^2 = \frac{1}{3}$. Computing the final expression yields $a^2 + \frac{1}{b^2} = 2 + 3 = 5$.

The locus of the midpoint of a line segment joining a fixed point and a point on a curve is found by expressing the variable point in parametric form and using the midpoint formula to relate the coordinates of the locus to the parameter. Given the ellipse $x^2 + 2y^2 = 4$, we can represent any point on it as $(2 \cos \theta, \sqrt{2} \sin \theta)$. Using the midpoint formula for the segment connecting this point and $(4, 3)$, the coordinates $(h, k)$ of the midpoint satisfy $h = \frac{2 \cos \theta + 4}{2}$ and $k = \frac{\sqrt{2} \sin \theta + 3}{2}$, which simplifies to $\cos \theta = h - 2$ and $\sin \theta = \frac{2k - 3}{\sqrt{2}}$. Substituting these into the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$ results in $(h - 2)^2 + \frac{(2k - 3)^2}{2} = 1$, which further rearranges to $(h - 2)^2 + \frac{(k - 1.5)^2}{1/2} = 1$. Comparing this to the standard ellipse equation $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, we identify $a^2 = 1$ and $b^2 = \frac{1}{2}$. The eccentricity is then determined by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$, yielding $e = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$.

A straight line with slope $m$ represents a common tangent if its vertical intercept $c$ satisfies the tangency condition for both curves simultaneously. For the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$, the condition for the line $y = mx + c$ to be tangent is $c^2 = 16m^2 + 9$. For the circle $x^2 + y^2 = 12$, the line must be at a distance equal to the radius $\sqrt{12}$ from the origin, expressed as $c^2 = 12(m^2 + 1)$. Setting these two values of $c^2$ equal to each other leads to the equation $16m^2 + 9 = 12m^2 + 12$. Simplifying this by rearranging terms results in $4m^2 = 3$. Multiplying both sides of this equality by $3$ directly produces the required value of $12m^2 = 9$.

Completing the square for the $x$ and $y$ terms in the equation $x^2+4y^2+2x+8y-\lambda=0$ transforms it into $(x^2+2x+1) + 4(y^2+2y+1) = \lambda+1+4$, which results in $(x+1)^2 + 4(y+1)^2 = \lambda+5$. Dividing both sides by $\lambda+5$ puts the ellipse in the standard form:

$\frac{(x+1)^2}{\lambda+5} + \frac{(y+1)^2}{\frac{\lambda+5}{4}} = 1$

This identifies $a^2 = \lambda+5$ and $b^2 = \frac{\lambda+5}{4}$. The formula for the length of the latus rectum is $\frac{2b^2}{a}$, and setting this equal to $4$ leads to:

$\frac{2(\frac{\lambda+5}{4})}{\sqrt{\lambda+5}} = 4$

This simplifies to $\frac{\sqrt{\lambda+5}}{2} = 4$, which implies $\sqrt{\lambda+5} = 8$, so $\lambda+5 = 64$ and $\lambda = 59$. The major axis length $l$ is defined as $2a$, or $2\sqrt{\lambda+5}$, giving $l = 2(8) = 16$. Finally, adding these values results in $\lambda+l = 59+16 = 75$.

The angle between two tangents drawn from an external point to an ellipse is determined by the slopes of these tangents. For the ellipse $2x^2 + 3y^2 = 5$, a line with slope $m$ passing through the point $(1, 3)$ follows the equation $y - 3 = m(x - 1)$, which simplifies to $y = mx - m + 3$. By applying the condition for tangency $y = mx \pm \sqrt{\frac{5}{2}m^2 + \frac{5}{3}}$, we set the intercept $3 - m$ equal to $\pm \sqrt{\frac{5}{2}m^2 + \frac{5}{3}}$. Squaring both sides yields the quadratic equation $\frac{3}{2}m^2 + 6m - \frac{22}{3} = 0$, which simplifies to $9m^2 + 36m - 44 = 0$.

The slopes $m_1$ and $m_2$ of the two tangents are the roots of this quadratic equation, providing the relationships $m_1 + m_2 = -4$ and $m_1m_2 = -\frac{44}{9}$. The difference between the slopes, $|m_1 - m_2|$, is found using the identity $\sqrt{(m_1 + m_2)^2 - 4m_1m_2}$, which evaluates to $\sqrt{16 - 4(-\frac{44}{9})} = \sqrt{\frac{320}{9}} = \frac{8\sqrt{5}}{3}$. The tangent of the acute angle $\alpha$ is calculated using the formula $\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$. Substituting the values results in $\tan \alpha = \left| \frac{8\sqrt{5}/3}{1 - 44/9} \right| = \left| \frac{8\sqrt{5}/3}{-35/9} \right| = \frac{24\sqrt{5}}{35}$, which simplifies to $\frac{24}{7\sqrt{5}}$.

The equation of a tangent to the parabola $y^2=4x$ is expressed as $y=mx+1/m$, where $m$ denotes the slope. Since this tangent passes through the point $(\alpha, \beta)$, substituting these coordinates into the equation yields the quadratic relationship $\alpha m^2 - \beta m + 1 = 0$. Given that the problem specifies two tangents with slopes in a ratio of $1:4$, we let the roots of this quadratic be $m_1$ and $4m_1$. Applying Vieta's formulas, the sum of the roots is $5m_1 = \beta/\alpha$ and the product of the roots is $4m_1^2 = 1/\alpha$. Squaring the expression for the sum gives $25m_1^2 = \beta^2/\alpha^2$, and substituting $m_1^2 = 1/(4\alpha)$ into this equation simplifies to $25/(4\alpha) = \beta^2/\alpha^2$, leading to the condition $4\beta^2 = 25\alpha$.

Substituting this result into the equation of the ellipse $25\alpha^2 + 4\beta^2 = 1$ gives $25\alpha^2 + 25\alpha = 1$, or equivalently $25(\alpha^2+\alpha)=1$. To determine the value of the target expression $(10\alpha+5)^2+(16\beta^2+50)^2$, we first rewrite the terms as $5(2\alpha+1)$ and $4(4\beta^2)+50$. Since $4\beta^2=25\alpha$, the second term becomes $4(25\alpha)+50$, which simplifies to $100\alpha+50$ or $50(2\alpha+1)$. Squaring these expressions yields $25(2\alpha+1)^2$ and $2500(2\alpha+1)^2$, respectively. Their sum is $2525(2\alpha+1)^2$, which expands to $2525(4\alpha^2+4\alpha+1)$. Because $\alpha^2+\alpha=1/25$, it follows that $4\alpha^2+4\alpha=4/25$. Substituting this into the expanded sum results in $2525(4/25+1)$, which is $2525 \cdot (29/25)$, equating to $101 \cdot 29$ or $2929$.

The intersection points of the line $y=x+1$ and the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ share $x$-coordinates that satisfy the quadratic equation found by substituting the linear expression into the ellipse equation. Replacing $y$ with $x+1$ in the ellipse equation yields $\frac{x^2}{4}+\frac{(x+1)^2}{2}=1$, which simplifies by multiplying through by 4 to $x^2+2(x^2+2x+1)=4$, or $3x^2+4x-2=0$.

If $\alpha$ and $\beta$ represent the roots of this quadratic, Vieta's formulas provide the sum $\alpha+\beta=-\frac{4}{3}$ and the product $\alpha\beta=-\frac{2}{3}$. The points $P(\alpha, \alpha+1)$ and $Q(\beta, \beta+1)$ define a segment $PQ$ with a squared length of $(\alpha-\beta)^2 + ((\alpha+1)-(\beta+1))^2 = 2(\alpha-\beta)^2$. Since this segment is the diameter of the circle, the radius $r$ relates to the diameter as $r = \frac{PQ}{2}$, implying $r^2 = \frac{2(\alpha-\beta)^2}{4} = \frac{(\alpha-\beta)^2}{2}$.

Applying the identity $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$, the squared difference is $\left(-\frac{4}{3}\right)^2 - 4\left(-\frac{2}{3}\right) = \frac{16}{9} + \frac{8}{3} = \frac{40}{9}$. The final required value is $(3r)^2 = 9r^2 = 9 \left( \frac{40/9}{2} \right) = 20$.

Completing the square for the $x$ terms in $x^{2}+9 y^{2}-4 x+3=0$ allows us to define the boundaries of the variables by grouping the quadratic terms and moving the constants to result in $(x-2)^{2}+9 y^{2}=1$. Writing this in standard form as $\frac{(x-2)^{2}}{1^{2}}+\frac{y^{2}}{(1/3)^{2}}=1$ reveals that the equation describes an ellipse centered at $(2, 0)$ with a horizontal extent of 1 unit and a vertical extent of $\frac{1}{3}$ units. Evaluating these geometric bounds confirms that $x$ lies in the interval $[1, 3]$ and $y$ falls within the range $[-\frac{1}{3}, \frac{1}{3}]$.

The equation of a tangent line to an ellipse can be expressed in the parametric form $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$, which corresponds to the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Dividing the given line equation $12x \cos \theta + 5y \sin \theta = 60$ by 60 transforms it into the standard form $\frac{x \cos \theta}{5} + \frac{y \sin \theta}{12} = 1$. Comparing this expression with the tangent formula identifies the parameters $a=5$ and $b=12$, confirming that the line is tangent to the ellipse $\frac{x^2}{25} + \frac{y^2}{144} = 1$, which simplifies to $144x^2 + 25y^2 = 3600$.

To find the slope of the common tangent line, we first express both given curves in their standard forms. The ellipse $4x^2 + 9y^2 = 36$ can be rewritten as $\frac{x^2}{9} + \frac{y^2}{4} = 1$, and the circle $(2x)^2 + (2y)^2 = 31$ simplifies to $x^2 + y^2 = \frac{31}{4}$. The equation of a line tangent to the ellipse with slope $m$ is given by $y = mx \pm \sqrt{9m^2 + 4}$, while the tangent to the circle is given by $y = mx \pm \sqrt{\frac{31}{4}(1 + m^2)}$. Since the line $L$ is a common tangent, the constant terms under the radicals must be equivalent, leading to the condition $9m^2 + 4 = \frac{31}{4}(1 + m^2)$. Multiplying the entire equation by 4 yields $36m^2 + 16 = 31 + 31m^2$, which simplifies to $5m^2 = 15$, confirming that the square of the slope is 3.