Ellipse – JEE Main MathematicsPractice Questions & PYQs

The semi-major axis length $a$ and semi-minor axis length $b$ are determined by the ellipse's eccentricity $e = \frac{1}{2}$ and foci at $(\pm 2, 0)$. Given that the distance to the foci is $ae = 2$, substituting the eccentricity yields $a(\frac{1}{2}) = 2$, so $a = 4$. The value of $b$ follows from the relation $b^2 = a^2(1 - e^2)$, resulting in $b^2 = 16(1 - \frac{1}{4}) = 12$, which gives $b = 2\sqrt{3}$. The minimum enclosing circle of an ellipse has a radius equal to the semi-major axis $a = 4$, centered at the origin.

Since the side $Q R$ is parallel to the major axis and contains the intersection point with the negative $y$-axis, which is $(0, -b)$, its position is defined by the line $y = -2\sqrt{3}$. The length of this base is given as $2a = 8$. To maximize the area of triangle $P Q R$, vertex $P$ must be positioned at the highest point of the enclosing circle, where $y = 4$. The altitude of the triangle is then the vertical distance from this peak to the base $y = -2\sqrt{3}$, calculating to $4 + 2\sqrt{3}$. Applying the formula $\frac{1}{2} \times \text{base} \times \text{height}$ results in an area of $\frac{1}{2} \times 8 \times (4 + 2\sqrt{3})$, which simplifies to $8(2 + \sqrt{3})$.

The eccentricity $e$ of an ellipse is determined by its semi-axes, and here it is given as the minimum value of the quadratic function $f(t)=t^2+t+\frac{11}{12}$. By finding the vertex of this parabola at $t=-\frac{1}{2}$, we calculate $e = (-\frac{1}{2})^2 - \frac{1}{2} + \frac{11}{12} = \frac{3-6+11}{12} = \frac{2}{3}$. Using the standard relation $e^2 = 1-\frac{b^2}{a^2}$, we substitute $e=\frac{2}{3}$ to obtain $\frac{4}{9} = 1-\frac{b^2}{a^2}$, which leads to the ratio $\frac{b^2}{a^2} = \frac{5}{9}$, or $b^2 = \frac{5a^2}{9}$.

The length of the latus rectum of this ellipse is given as $10$, and applying the formula $\frac{2b^2}{a} = 10$, we deduce $b^2 = 5a$. Combining these two expressions for $b^2$, we equate $5a = \frac{5a^2}{9}$, which simplifies to $a=9$ since the semi-major axis must be positive. This result leads to $b^2 = 5(9) = 45$, and the final required sum $a^2+b^2$ is $81+45=126$.

To find the equation of a chord of an ellipse when the midpoint is known, we utilize the standard formula $T=S_1$. For an ellipse defined by the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the chord passing through a midpoint $(x_1, y_1)$ is expressed as $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}$. Substituting the given parameters, where $a^2=9$, $b^2=4$, and the midpoint is $\left(\frac{5}{2}, \frac{1}{2}\right)$, the equation becomes $\frac{x(5/2)}{9}+\frac{y(1/2)}{4}=\frac{(5/2)^2}{9}+\frac{(1/2)^2}{4}$.

Simplifying the right side involves calculating $\frac{25}{36}+\frac{1}{16}$, which results in $\frac{109}{144}$ after finding a common denominator. This gives us the equation $\frac{5x}{18}+\frac{y}{8}=\frac{109}{144}$. By multiplying the entire equation by $144$ to clear the denominators, we arrive at $40x+18y=109$. Comparing this to the required form $\alpha x+\beta y=109$, we identify $\alpha=40$ and $\beta=18$, which makes their sum $58$.

For the ellipse $E_1$ with the equation $\frac{x^2}{9} + \frac{y^2}{4} = 1$, the semi-major axis is $a_1 = 3$ and the semi-minor axis is $b_1 = 2$, resulting in an initial area $A_1 = \pi a_1 b_1 = 6\pi$. Because all ellipses share the same eccentricity, the ratio of the minor axis to the major axis remains invariant, specifically $\frac{b_i}{a_i} = \frac{2}{3}$ for every ellipse in the sequence.

Since the major axis of each subsequent ellipse is equal to the minor axis of the preceding one, we have the relation $a_{i+1} = b_i$. Combining this with the constant ratio $\frac{b_{i+1}}{a_{i+1}} = \frac{2}{3}$, it follows that $b_{i+1} = \frac{2}{3} a_{i+1} = \frac{2}{3} b_i$. The area of the $(i+1)$-th ellipse is $A_{i+1} = \pi a_{i+1} b_{i+1} = \pi b_i \left( \frac{2}{3} b_i \right)$, which simplifies to $A_{i+1} = \frac{2}{3} \pi b_i^2$. Comparing this to $A_i = \pi a_i b_i = \frac{2}{3} \pi a_i^2$ (since $b_i = \frac{2}{3} a_i$), the ratio between consecutive areas is $\frac{A_{i+1}}{A_i} = \frac{b_i^2}{a_i^2} = (\frac{2}{3})^2 = \frac{4}{9}$.

These areas form an infinite geometric series with a first term of $6\pi$ and a common ratio of $\frac{4}{9}$. The sum of this series is given by $\frac{6\pi}{1 - 4/9} = \frac{6\pi}{5/9} = \frac{54\pi}{5}$. Multiplying this sum by the factor $\frac{5}{\pi}$ yields $\frac{5}{\pi} \cdot \frac{54\pi}{5} = 54$.

The equation of a chord of an ellipse with a known midpoint $(h, k)$ is given by the relation $T=S_1$, where $T$ represents the tangent equation form evaluated at the midpoint and $S_1$ is the value of the ellipse equation at that same point. Applying this to the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ at the midpoint $(1, \frac{1}{2})$ produces $\frac{x(1)}{4} + \frac{y(1/2)}{2} = \frac{1^2}{4} + \frac{(1/2)^2}{2}$, which simplifies to the linear equation $2x+2y=3$. Expressing $y$ as $y=\frac{3-2x}{2}$ and substituting this into the original ellipse equation gives $\frac{x^2}{4}+\frac{(\frac{3-2x}{2})^2}{2}=1$, identifying the intersection points $x=\frac{12 \pm \sqrt{120}}{12}$ and $y=\frac{1}{2} \mp \frac{\sqrt{120}}{12}$. Calculating the distance between these two points determines the chord length to be $\frac{2}{3}\sqrt{15}$.

The minor axis length of an ellipse is $2b$, and the distance between the foci is $2ae$, where $e$ is the eccentricity. Given that the minor axis is one-fourth of the focal distance, the equation relating these terms is:

$2b = \frac{1}{4}(2ae)$

Simplifying this equation leads to $b = \frac{ae}{4}$, and squaring both sides yields $b^2 = \frac{a^2e^2}{16}$. Substituting the standard identity $b^2 = a^2(1 - e^2)$ into this expression produces $a^2(1 - e^2) = \frac{a^2e^2}{16}$. Dividing the entire equation by $a^2$ results in $1 - e^2 = \frac{e^2}{16}$. Combining the $e^2$ terms gives $1 = \frac{17e^2}{16}$. Solving for the eccentricity results in:

$e = \frac{4}{\sqrt{17}}$

The equation of a chord bisected at a specific point on an ellipse is defined by the relation $T = S_1$, where $T$ represents the expression obtained by replacing $x^2$ with $x x_1$ and $y^2$ with $y y_1$ in the ellipse equation, and $S_1$ is the ellipse equation evaluated at the coordinates of that mid-point. For the ellipse defined by $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and the mid-point $(3, 1)$, we determine $T$ as $\frac{3x}{25} + \frac{y}{16} - 1$ and $S_1$ as the value of the ellipse equation at $(3, 1)$, which is $\frac{3^2}{25} + \frac{1^2}{16} - 1$.

Setting these two expressions equal causes the constant term of $-1$ to cancel from both sides, simplifying the relation to $\frac{3x}{25} + \frac{y}{16} = \frac{9}{25} + \frac{1}{16}$. Multiplying the entire equation by the common denominator of $400$ clears the fractions, resulting in $16(3x) + 25(y) = 9(16) + 25$. Completing the arithmetic on the right side, $144 + 25$, yields the final equation:
$48x + 25y = 169$

The product of the focal distances for any point $(x_1, y_1)$ on an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is equal to $a^2 - e^2x_1^2$. For the point $\left(\sqrt{3}, \;\frac{1}{2}\right)$, this product equates to $a^2 - 3e^2 = \frac{7}{4}$, which allows us to express $a^2 = 3e^2 + \frac{7}{4}$. Since the point lies on the ellipse, it must satisfy the equation $\frac{3}{a^2} + \frac{1}{4b^2} = 1$, and by substituting the standard identity $b^2 = a^2(1-e^2)$, this transforms into $\frac{3}{a^2} + \frac{1}{4a^2(1-e^2)} = 1$.

Multiplying this expression by $4a^2(1-e^2)$ yields $12(1-e^2) + 1 = 4a^2(1-e^2)$, which simplifies further by substituting our expression for $a^2$ into $13 - 12e^2 = 4(3e^2 + \frac{7}{4})(1-e^2)$. Simplifying this algebraic structure results in the equation
$12e^4 - 17e^2 + 6 = 0$
Factoring the polynomial in terms of $e^2$ gives $(3e^2 - 2)(4e^2 - 3) = 0$, identifying the eccentricities as $e = \sqrt{\frac{2}{3}}$ and $e = \sqrt{\frac{3}{4}}$. The absolute difference between these two values is calculated as $\left| \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{\sqrt{3}} \right|$, which simplifies to $\frac{3 - 2\sqrt{2}}{2\sqrt{3}}$.

The condition for the line $y=mx+c$ to be tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is defined by $c^2 = a^2m^2 + b^2$. Applying the ellipse parameters $a^2=16$ and $b^2=9$ with the line slope $m=1$, we determine $c^2 = 16(1)^2 + 9 = 25$, which yields $c = \pm 5$. The points of tangency $A$ and $B$ for the lines $y=x+5$ and $y=x-5$ are found to be $A(-\frac{16}{5}, \frac{9}{5})$ and $B(\frac{16}{5}, -\frac{9}{5})$.

Simultaneously, the line $y=x$ intersects the ellipse at the points $C(\frac{12}{5}, \frac{12}{5})$ and $D(-\frac{12}{5}, -\frac{12}{5})$. When these four points are considered as vertices in the order $ABCD$, the resulting segments intersect each other rather than forming the boundary of a simple polygon. Because the points do not form a conventional, non-intersecting quadrilateral, the area cannot be determined in the standard way, and none of the provided numerical options match the geometric outcome.

Combinatorial geometry dictates that the number of triangles and quadrilaterals that can be formed by joining the vertices of a polygon with $n$ sides is determined by the combinations $\binom{n}{3}$ and $\binom{n}{4}$, respectively. Given their sum $p+q = \binom{n}{3} + \binom{n}{4} = 126$, we apply Pascal's Identity, $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$, to simplify the expression to $\binom{n+1}{4} = 126$. Expanding this binomial coefficient as $\frac{(n+1)n(n-1)(n-2)}{24} = 126$ leads to the product of four consecutive integers $(n+1)n(n-1)(n-2) = 3024$. Since $9 \times 8 \times 7 \times 6 = 3024$, we identify $n+1 = 9$, which gives $n = 8$.

With the number of sides $n = 8$, the ellipse equation is written as $\frac{x^2}{16} + \frac{y^2}{8} = 1$. Comparing this to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we have $a^2 = 16$ and $b^2 = 8$. The eccentricity of the ellipse is calculated using the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$, which becomes $e = \sqrt{1 - \frac{8}{16}} = \sqrt{1 - \frac{1}{2}}$, resulting in $e = \frac{1}{\sqrt{2}}$.

Since both the ellipse and the hyperbola share a focal distance of $2\sqrt{3}$, their respective focal parameters $ae$ and $Ae'$ must equal $\sqrt{3}$. The ratio of their eccentricities $\frac{e}{e'} = \frac{1}{3}$ implies an inverse relationship with their semi-axes, expressed as $\frac{A}{a} = \frac{1}{3}$, which simplifies to $a = 3A$. By substituting this into the condition $a - A = 2$, we solve for the semi-major axes to find $2A = 2$, yielding $A = 1$ and $a = 3$.

Next, we determine the eccentricities using the relations $3e = \sqrt{3}$ and $1e' = \sqrt{3}$, which result in $e = \frac{1}{\sqrt{3}}$ and $e' = \sqrt{3}$. The square of the ellipse's semi-minor axis is calculated via $b^2 = a^2(1 - e^2) = 9(1 - \frac{1}{3}) = 6$, while the square of the hyperbola's conjugate semi-axis is $B^2 = A^2((e')^2 - 1) = 1(3 - 1) = 2$. Calculating the latus rectum lengths using $\frac{2b^2}{a}$ and $\frac{2B^2}{A}$ provides values of 4 and 4 respectively, which combine to a total sum of 8.

The geometry of this problem is best analyzed by using the parametric form of a line passing through the point $P(\sqrt{5}, \sqrt{5})$, expressed as $x = \sqrt{5} + r \cos \theta$ and $y = \sqrt{5} + r \sin \theta$. Substituting these into the ellipse equation $25x^2 + 36y^2 = 900$ generates a quadratic equation in $r$, where the absolute product of the roots $|r_1 r_2|$ represents the magnitude of the product of the segment lengths $PA$ and $PB$. This product is equal to $\frac{595}{25 \cos^2 \theta + 36 \sin^2 \theta}$, which reaches its maximum value when the denominator is minimized, occurring at $\theta = 0$. This alignment defines a horizontal line $y = \sqrt{5}$, and intersecting this line with the ellipse yields $\frac{x^2}{36} + \frac{5}{25} = 1$, or $x = \pm \frac{12}{\sqrt{5}}$. With the intersection points being $(\frac{12}{\sqrt{5}}, \sqrt{5})$ and $(-\frac{12}{\sqrt{5}}, \sqrt{5})$, the squared distances are calculated as $PA^2 = (\frac{12}{\sqrt{5}} - \sqrt{5})^2 = \frac{49}{5}$ and $PB^2 = (-\frac{12}{\sqrt{5}} - \sqrt{5})^2 = \frac{289}{5}$. Summing these values gives $\frac{338}{5}$, and multiplying this sum by 5 yields 338.

The points $A(10, 0)$ and $B(0, \frac{50}{7})$ represent the intercepts of the line $5x + 7y = 50$, and application of the section formula for an internal ratio of $7:3$ locates $P$ at $(3, 5)$. Since the perpendicular from the focus $S$ to the x-axis passes through $P$, the x-coordinate of the focus must be $3$, which identifies $ae = 3$.

The directrix equation $3x - 25 = 0$ corresponds to $\frac{a}{e} = \frac{25}{3}$. Combining the expressions $ae = 3$ and $\frac{a}{e} = \frac{25}{3}$ leads to $a^2 = 25$, so $a = 5$, and $e^2 = \frac{9}{25}$. With $b^2 = a^2(1 - e^2) = 16$, the length of the latus rectum follows as $\frac{2b^2}{a} = \frac{2(16)}{5} = \frac{32}{5}$.

The length of the minor axis of an ellipse is represented by $2b$, while the distance between the foci is $2ae$, where $a$ denotes the semi-major axis and $e$ is the eccentricity. Given that the minor axis length equals half the focal distance, the relationship $2b = ae$ applies, which simplifies to the ratio $\frac{b}{a} = \frac{e}{2}$. By squaring this ratio and substituting it into the standard geometric identity $\frac{b^2}{a^2} = 1 - e^2$, the equation becomes $\frac{e^2}{4} = 1 - e^2$. Combining the terms involving $e^2$ results in $\frac{5e^2}{4} = 1$, which reveals that the eccentricity is $e = \frac{2}{\sqrt{5}}$.

The parametric coordinates of any point $P$ on the ellipse defined by $\frac{x^2}{9} + \frac{y^2}{4} = 1$ can be expressed as $(3 \cos \theta, 2 \sin \theta)$. Since the line passing through $P$ is parallel to the $y$-axis, it intersects the circle $x^2 + y^2 = 9$ at a point $Q$ that shares the same $x$-coordinate, $3 \cos \theta$. Substituting this into the circle's equation gives $9 \cos^2 \theta + y_Q^2 = 9$, which simplifies to $y_Q^2 = 9 \sin^2 \theta$. Given that $P$ and $Q$ lie on the same side of the $x$-axis, the $y$-coordinate of $Q$ is $3 \sin \theta$, establishing $Q$ as $(3 \cos \theta, 3 \sin \theta)$.

Point $R(x, y)$ divides the line segment $PQ$ in the ratio $4 : 3$. Applying the section formula for internal division, the coordinates of $R$ are calculated as $x = \frac{4(3 \cos \theta) + 3(3 \cos \theta)}{7} = 3 \cos \theta$ and $y = \frac{4(3 \sin \theta) + 3(2 \sin \theta)}{7} = \frac{18}{7} \sin \theta$. By isolating the trigonometric functions, we obtain $\cos \theta = \frac{x}{3}$ and $\sin \theta = \frac{7y}{18}$. Substituting these into the Pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$ yields the equation for the locus of $R$:
$\frac{x^2}{9} + \frac{49 y^2}{324} = 1$

This locus is an ellipse where the semi-major axis $a$ equals $3$ and the semi-minor axis $b$ equals $\frac{18}{7}$. The eccentricity $e$ is determined using the relation $e = \sqrt{1 - \frac{b^2}{a^2}}$, which results in $\sqrt{1 - \frac{324/49}{9}}$. Simplifying this expression leads to $\sqrt{1 - \frac{36}{49}}$, which equals $\sqrt{\frac{13}{49}}$, or $\frac{\sqrt{13}}{7}$.

The center of a circle with a diameter segment on a specific line must lie on that line, allowing us to find the constant $k$ by substituting the coordinates of the circle's center into the equation $2x+3y-k=0$. The circle $x^2+y^2-3x-2y=0$ has its center at $(\frac{3}{2}, 1)$ by examining the coefficients, and substituting these coordinates into the line equation yields $2(\frac{3}{2}) + 3(1) - k = 0$, which results in $k=6$.

With $k$ determined, the ellipse equation becomes $x^2+9y^2=36$, which simplifies to the standard form $\frac{x^2}{6^2} + \frac{y^2}{2^2} = 1$. The length of the latus rectum for an ellipse is calculated using $\frac{2b^2}{a}$, resulting in $\frac{2(2^2)}{6} = \frac{8}{6} = \frac{4}{3}$, where $m=4$ and $n=3$. Substituting these values into the required expression gives $2m+n = 2(4)+3 = 11$.

The equation of a chord of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with a given midpoint $(x_0, y_0)$ is determined by the relation $T = S_1$, where $T = \frac{x x_0}{a^2} + \frac{y y_0}{b^2}$ and $S_1 = \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2}$. For the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ with midpoint $\left(1, \frac{2}{5}\right)$, this becomes $\frac{x}{25} + \frac{y(2/5)}{16} = \frac{1^2}{25} + \frac{(2/5)^2}{16}$. Simplifying the right side gives $\frac{1}{25} + \frac{4/25}{16} = \frac{1}{25} + \frac{1}{100} = \frac{5}{100} = \frac{1}{20}$, leading to the linear equation $\frac{x}{25} + \frac{y}{40} = \frac{1}{20}$, which further simplifies to $y = \frac{10 - 8x}{5}$.

Substituting this linear expression into the original ellipse equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$ results in the quadratic equation $4x^2 - 8x - 15 = 0$. Solving for $x$ yields two roots, $x = 1 \pm \frac{\sqrt{304}}{8}$, representing the x-coordinates of the chord's endpoints. Using these values to find the corresponding y-coordinates and applying the distance formula between the two points, the final length of the chord is determined to be $\frac{\sqrt{1691}}{5}$.

Calculating the parameters $a$ and $b$ requires evaluating the composite functions $f(g(10))$ and $g(f(3))$ using the definitions $f(x)=x^2+9$ and $g(x)=\frac{x}{x-9}$. Substituting $10$ into $g(x)$ gives $g(10) = \frac{10}{10-9} = 10$, which leads to $a = f(10) = 10^2 + 9 = 109$. Similarly, evaluating $f(3) = 3^2+9 = 18$ allows us to find $b = g(18) = \frac{18}{18-9} = 2$. With these values, the ellipse is defined by the equation $\frac{x^2}{109} + \frac{y^2}{2} = 1$.

For this ellipse, the squared eccentricity $e^2$ is determined by $1 - \frac{b}{a} = 1 - \frac{2}{109} = \frac{107}{109}$. The length of the latus rectum $l$ is calculated using the formula $\frac{2B^2}{A}$, where $A^2=109$ and $B^2=2$, yielding $l = \frac{2 \times 2}{\sqrt{109}} = \frac{4}{\sqrt{109}}$. Squaring this length gives $l^2 = \frac{16}{109}$. Combining these results into the expression $8e^2 + l^2$ yields $8\left(\frac{107}{109}\right) + \frac{16}{109}$, which simplifies to $\frac{856 + 16}{109} = \frac{872}{109} = 8$.

For a circle centered at $(2,0)$ to be inscribed in the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$, it must be tangent to the ellipse at a point where the normal to the ellipse passes through the center of the circle. The equation of the normal to this ellipse at any point $(6 \cos \theta, 4 \sin \theta)$ is given by $3x \sec \theta - 2y \csc \theta = 10$. Substituting the circle's center $(2,0)$ into this normal equation, we get $3(2) \sec \theta = 10$, which simplifies to $\sec \theta = \frac{5}{3}$, or $\cos \theta = \frac{3}{5}$. Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, we find $\sin \theta = \frac{4}{5}$, identifying the point of tangency as $(\frac{18}{5}, \frac{16}{5})$.

The squared radius $r^2$ of the circle is the squared distance between the center $(2,0)$ and the point of tangency $(\frac{18}{5}, \frac{16}{5})$. Calculating this, we have $r^2 = (\frac{18}{5}-2)^2 + (\frac{16}{5}-0)^2 = (\frac{8}{5})^2 + (\frac{16}{5})^2 = \frac{64+256}{25} = \frac{320}{25} = \frac{64}{5}$. The equation of the circle is therefore $(x-2)^2 + y^2 = \frac{64}{5}$.

Since the point $(1, \alpha)$ lies on the circle, we substitute these coordinates into the equation to get $(1-2)^2 + \alpha^2 = \frac{64}{5}$. This simplifies to $1 + \alpha^2 = \frac{64}{5}$, which means $\alpha^2 = \frac{59}{5}$. Finally, multiplying this result by 10, we obtain $10 \alpha^2 = 10 \left( \frac{59}{5} \right) = 118$.

The ellipse defined by $k x^2 + k^2 y^2 = 1$ can be rewritten in standard form as $\frac{x^2}{1/k} + \frac{y^2}{1/k^2} = 1$, which identifies the semi-axes lengths as $a = \frac{1}{\sqrt{k}}$ and $b = \frac{1}{k}$. The four chords connecting the endpoints of these axes are lines represented by the equation $\sqrt{k} x + k y = 1$ in the first quadrant. The radius $r_k$ of the circle $C_k$ centered at the origin is equivalent to the perpendicular distance from the origin to any one of these lines, given by $r_k = \frac{|-1|}{\sqrt{(\sqrt{k})^2 + k^2}} = \frac{1}{\sqrt{k + k^2}}$. Squaring the reciprocal of this radius yields $\frac{1}{r_k^2} = k + k^2$. Summing these terms from $k=1$ to $20$ requires calculating $\sum_{k=1}^{20} k + \sum_{k=1}^{20} k^2$. Applying the standard summation formulas results in $\frac{20 \times 21}{2} + \frac{20 \times 21 \times 41}{6} = 210 + 2870$, which equals $3080$.