Ellipse – JEE Main MathematicsPractice Questions & PYQs

A line $y = mx + c$ functions as a common tangent to two curves only if it simultaneously satisfies the tangency condition for each. For the parabola $y^2 = 16\sqrt{3}x$, the condition for a line with slope $m$ to be tangent is $c = \frac{4\sqrt{3}}{m}$, whereas for the ellipse $2x^2 + y^2 = 4$, expressed in standard form as $\frac{x^2}{2} + \frac{y^2}{4} = 1$, the condition is $c^2 = 2m^2 + 4$. By equating the squared constant terms, specifically $(\frac{4\sqrt{3}}{m})^2 = 2m^2 + 4$, we obtain the relationship $\frac{48}{m^2} = 2m^2 + 4$. Multiplying by $m^2$ yields $48 = 2m^4 + 4m^2$, which simplifies to $m^4 + 2m^2 = 24$, confirming the second statement. Solving this biquadratic equation results in $m^2 = 4$, meaning $m = \pm 2$. Substituting $m = 2$ back into the tangent form $y = mx + \frac{4\sqrt{3}}{m}$ gives the equation $y = 2x + 2\sqrt{3}$, which proves the first statement is true. Because the condition derived in the second statement serves as the necessary logical basis to obtain the specific equation in the first statement, it acts as the correct explanation.

The shape of the ellipse is established by determining its semi-axes from the provided circle equations. The circle $(x-1)^2+y^2=1$ has a radius of 1 and therefore a diameter of 2, which defines the length of the semi-minor axis. Conversely, the circle $x^2+(y-2)^2=4$ has a radius of 2 and a diameter of 4, which identifies the length of the semi-major axis.

Substituting these dimensions into the standard ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where the semi-major axis $a=4$ and the semi-minor axis $b=2$, results in $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Multiplying the entire equation by 16 to eliminate the denominators produces the final equation $x^2+4y^2=16$.

The standard equation for an ellipse centered at the origin with its major and minor axes along the coordinate axes is represented by:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
Given the eccentricity $e = \sqrt{\frac{2}{5}}$, we use the relation $e^2 = 1 - \frac{b^2}{a^2}$ to determine that $b^2 = a^2(1 - \frac{2}{5})$, which simplifies to $b^2 = \frac{3}{5}a^2$.

Since the ellipse passes through the point $(-3, 1)$, substituting these coordinates into the standard equation results in the condition:
$\frac{9}{a^2} + \frac{1}{b^2} = 1$
Replacing $b^2$ with $\frac{3}{5}a^2$ gives $\frac{9}{a^2} + \frac{5}{3a^2} = 1$. Solving this for $a^2$ by finding a common denominator yields $\frac{27 + 5}{3a^2} = 1$, which simplifies to $3a^2 = 32$, or $a^2 = \frac{32}{3}$. Consequently, $b^2 = \frac{3}{5}(\frac{32}{3}) = \frac{32}{5}$. Substituting these values back into the ellipse equation produces $\frac{3x^2}{32} + \frac{5y^2}{32} = 1$, which simplifies to:
$3x^2 + 5y^2 = 32$

The inner ellipse $x^2 + 4y^2 = 4$ simplifies to $\frac{x^2}{4} + \frac{y^2}{1} = 1$, identifying its vertices at $(\pm 2, 0)$ and $(0, \pm 1)$. A bounding rectangle aligned with the coordinate axes must touch these vertices, which places the corner point $(2, 1)$ on the perimeter of the outer ellipse.

The outer ellipse is defined by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Since it passes through the point $(4, 0)$, substituting these coordinates gives $\frac{16}{a^2} = 1$, establishing that $a^2 = 16$. Using the corner point $(2, 1)$ to satisfy this equation results in $\frac{4}{16} + \frac{1}{b^2} = 1$, which simplifies to $\frac{1}{4} + \frac{1}{b^2} = 1$. Solving for the remaining parameter yields $\frac{1}{b^2} = \frac{3}{4}$, or $b^2 = \frac{4}{3}$. Substituting these values into the standard form produces the final equation:

$x^2 + 12y^2 = 16$

The fundamental relationship connecting the distance between an ellipse's focus and its directrix to its geometric parameters is defined by the equation $\frac{a}{e} - ae = d$, where $a$ is the semi-major axis, $e$ is the eccentricity, and $d$ is the perpendicular distance from the focus to the directrix. Positioning the focus at the origin and the directrix at $x=4$ establishes that the distance $d$ is 4. Given an eccentricity of $\frac{1}{2}$, substituting these known values into the equation yields $4 = \frac{a}{1/2} - a\left(\frac{1}{2}\right)$, which simplifies to $4 = 2a - \frac{a}{2}$. Combining the terms on the right side produces $4 = \frac{3a}{2}$, which leads directly to the result $a = \frac{8}{3}$.

The distance between the foci is $2ae = 6$, which implies $ae = 3$, and the minor axis is $2b = 8$, meaning $b = 4$. These dimensions are connected by the equation $b^2 = a^2 - (ae)^2$, where $a$ represents the semi-major axis. Substituting the known values into this relationship yields $16 = a^2 - 9$, which leads to $a^2 = 25$, so $a = 5$. The eccentricity is determined by the ratio $e = \frac{ae}{a}$, so dividing 3 by 5 provides the final result.

In an ellipse where the triangle formed by the two foci $F$ and $F^{\prime}$ and the end of the semi-minor axis $B$ is a right-angled triangle at $B$, the Pythagorean theorem dictates that $FB^{2} + F^{\prime}B^{2} = FF^{\prime 2}$. Since the distance from a focus to the end of the semi-minor axis is $\sqrt{a^{2}e^{2} + b^{2}}$ and the distance between the two foci is $2ae$, substituting these values produces the equation $2(a^{2}e^{2} + b^{2}) = (2ae)^{2}$, which simplifies to $a^{2}e^{2} + b^{2} = 2a^{2}e^{2}$, or simply $b^{2} = a^{2}e^{2}$. By substituting the fundamental identity $b^{2} = a^{2}(1 - e^{2})$ into this equation, we obtain $a^{2}e^{2} = a^{2}(1 - e^{2})$, which reduces to $2e^{2} = 1$ and results in the eccentricity $e = \frac{1}{\sqrt{2}}$.

The geometry of an ellipse is fundamentally determined by its semi-axes lengths, which are directly related to the eccentricity and the distance from the center to the directrix. For an ellipse centered at the origin, the directrix is defined by the equation $x = \frac{a}{e}$. Given the directrix is $x = 4$ and the eccentricity is $e = \frac{1}{2}$, we find the semi-major axis by calculating $a = 4 \times \frac{1}{2} = 2$.

To find the semi-minor axis $b$, we use the relationship $b^2 = a^2(1 - e^2)$. Substituting the values $a = 2$ and $e = \frac{1}{2}$ provides $b^2 = 2^2(1 - (\frac{1}{2})^2)$, which simplifies to $b^2 = 4(\frac{3}{4}) = 3$. Inserting $a^2 = 4$ and $b^2 = 3$ into the standard form equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ gives us $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Multiplying the entire equation by 12 clears the denominators to produce the final result:

$3x^2 + 4y^2 = 12$

For an ellipse defined by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the eccentricity $e$ follows the fundamental relationship $e^2 = 1 - \frac{b^2}{a^2}$. Given the eccentricity value of $\frac{1}{\sqrt{2}}$, squaring both sides reveals that $\frac{1}{2} = 1 - \frac{b^2}{a^2}$, which directly leads to the ratio $\frac{b^2}{a^2} = \frac{1}{2}$.

The hyperbola represented by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has an eccentricity $e_H$ defined by the equation
$e_H^2 = 1 + \frac{b^2}{a^2}$
Substituting the ratio $\frac{b^2}{a^2} = \frac{1}{2}$ derived from the ellipse into this expression results in $1 + \frac{1}{2}$, providing a final value of $\frac{3}{2}$. Because the square of the eccentricity of the hyperbola depends solely on the ratio of the squared axes of the ellipse, the additional information provided regarding the latus rectum is unnecessary for calculating this specific result.

The tangent line to the ellipse $\frac{x^2}{36}+\frac{y^2}{4}=1$ at the point $(3\sqrt{3},1)$ is given by $\frac{3\sqrt{3}x}{36} + \frac{1y}{4} = 1$, which simplifies to $\frac{x\sqrt{3}}{12} + \frac{y}{4} = 1$. Setting $x=0$ identifies the y-intercept $A(0, 4)$. The slope of this tangent is $-\frac{\sqrt{3}}{3}$, meaning the normal line has a slope of $\sqrt{3}$. Its equation $y - 1 = \sqrt{3}(x - 3\sqrt{3})$ simplifies to $y = \sqrt{3}x - 8$, which intersects the y-axis at $B(0, -8)$ when $x=0$.

A circle $C$ with $AB$ as its diameter has its center at the midpoint $(0, -2)$ and a radius of $6$, yielding the equation $x^2 + (y+2)^2 = 36$, or $x^2 + y^2 + 4y - 32 = 0$. The point $(\alpha, \beta)$ where the tangents at the intersection points $P$ and $Q$ of the line $x = 2\sqrt{5}$ meet is the pole of that line with respect to the circle. The equation of the polar line of $(\alpha, \beta)$ with respect to the circle is $x\alpha + y\beta + 2(y+\beta) - 32 = 0$, which can be rearranged to $x\alpha + y(\beta+2) + (2\beta-32) = 0$.

Comparing this polar equation to $x - 2\sqrt{5} = 0$, the ratios of their coefficients must be equal: $\frac{\alpha}{1} = \frac{\beta+2}{0} = \frac{2\beta-32}{-2\sqrt{5}}$. The middle term requires $\beta+2 = 0$, so $\beta = -2$. Substituting this value into the proportionality gives $\alpha = \frac{2(-2)-32}{-2\sqrt{5}} = \frac{-36}{-2\sqrt{5}} = \frac{18}{\sqrt{5}}$. Calculating the final expression leads to $\alpha^2 - \beta^2 = \left(\frac{18}{\sqrt{5}}\right)^2 - (-2)^2 = \frac{324}{5} - 4 = \frac{324-20}{5} = \frac{304}{5}$.

The defining properties of an ellipse establish a relationship between its directrix and focus governed by the eccentricity, specifically through the expressions $\frac{a}{e}=8$ and $ae=2$. Multiplying these equations yields $a^2=16$, which confirms $a=4$, and subsequently $e=\frac{1}{2}$ when substituted back into $ae=2$. With the semi-major axis $a=4$ and eccentricity $e=\frac{1}{2}$, the semi-minor axis is determined by $b^2=a^2(1-e^2)$, resulting in $b^2=16(1-\frac{1}{4})=12$, so the ellipse equation is $\frac{x^2}{16}+\frac{y^2}{12}=1$.

To identify point $P(x_1, y_1)$ on the ellipse where the tangent passes through $(0, 4\sqrt{3})$, the equation of the tangent $\frac{x x_1}{16} + \frac{y y_1}{12} = 1$ is employed. Substituting the point $(0, 4\sqrt{3})$ gives $\frac{4\sqrt{3}y_1}{12}=1$, simplifying to $y_1=\sqrt{3}$. Plugging this into the ellipse equation gives $\frac{x_1^2}{16} + \frac{3}{12} = 1$, which implies $x_1^2=12$, leading to $x_1=2\sqrt{3}$ for a point in the first quadrant. With $P(2\sqrt{3}, \sqrt{3})$, the tangent equation becomes $\frac{2\sqrt{3}x}{16} + \frac{\sqrt{3}y}{12} = 1$, which simplifies to $\frac{\sqrt{3}x}{8} + \frac{\sqrt{3}y}{12} = 1$.

The intersection with the $x$-axis at $Q$ occurs when $y=0$, leading to $\frac{\sqrt{3}x}{8}=1$ or $x=\frac{8}{\sqrt{3}}$, thus $Q(\frac{8}{\sqrt{3}}, 0)$. The square of the distance $PQ$ is $(2\sqrt{3}-\frac{8}{\sqrt{3}})^2 + (\sqrt{3}-0)^2 = (-\frac{2}{\sqrt{3}})^2 + 3 = \frac{4}{3} + 3 = \frac{13}{3}$. Finalizing the value, $(3PQ)^2 = 9 \times \frac{13}{3} = 39$.

A parabola's focus, vertex, and the intersection point of its axis and directrix are collinear, with the vertex serving as the midpoint of the segment connecting the focus and this intersection point. Given the directrix $2x + y - 6 = 0$ has a slope of $-2$, the axis of the parabola must have a slope of $1/2$. Since the axis passes through the vertex $(2, 3)$, its equation is $y - 3 = \frac{1}{2}(x - 2)$, which simplifies to $x - 2y + 4 = 0$. By solving the system formed by the directrix and axis equations, the intersection point of the axis and directrix is found to be $(1.6, 2.8)$. Using the midpoint property, where the vertex $(2, 3)$ is the average of the focus $(x, y)$ and the intersection point $(1.6, 2.8)$, the focus $S$ is calculated as $(2 \times 2 - 1.6, 2 \times 3 - 2.8) = (2.4, 3.2)$.

The ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through $(2.4, 3.2)$, leading to the equation $\frac{(2.4)^2}{a^2} + \frac{(3.2)^2}{b^2} = 1$. With an eccentricity $e = \frac{1}{\sqrt{2}}$, we have the relation $1 - \frac{b^2}{a^2} = \frac{1}{2}$, which simplifies to $a^2 = 2b^2$. Substituting this into the ellipse equation gives $\frac{5.76}{2b^2} + \frac{10.24}{b^2} = 1$, or $\frac{2.88 + 10.24}{b^2} = 1$, which determines $b^2 = 13.12 = \frac{328}{25}$. The square of the length of the latus rectum is given by $\left(\frac{2b^2}{a}\right)^2 = \frac{4b^4}{a^2}$. Because $a^2 = 2b^2$, this expression becomes $2b^2$, yielding $2 \times \frac{328}{25} = \frac{656}{25}$.

For the line segment joining the points of intersection of the parabola $y^{2}=4kx$ and the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ to be the latus rectum for both, their focal x-coordinates must coincide, establishing that $k = ae$. The length of the latus rectum for this parabola is $4k$, while the length for the ellipse is $\frac{2b^{2}}{a}$, and equating these two values gives $4k = \frac{2b^{2}}{a}$, which simplifies to $2a^{2}e = b^{2}$ after substituting $k = ae$. Since the semi-minor axis of an ellipse relates to the semi-major axis and eccentricity via $b^{2}=a^{2}(1-e^{2})$, replacing $b^{2}$ in the equation yields $2a^{2}e = a^{2}(1-e^{2})$, which reduces to $2e = 1-e^{2}$ after dividing both sides by $a^{2}$. Solving the resulting quadratic $e^{2}+2e-1=0$ for the positive eccentricity provides $e = \sqrt{2}-1$. Substituting this result into the expression $e^{2}+2\sqrt{2}$ leads to $(\sqrt{2}-1)^{2}+2\sqrt{2}$, which expands to $(2-2\sqrt{2}+1)+2\sqrt{2}$ and simplifies to 3.

Because $PQ$ and $RS$ are diameters of the ellipse $9x^2+4y^2=36$, they pass through the origin and are bisected by it. This means the lengths of the segments are $PQ = 2(OP)$ and $RS = 2(OR)$, allowing the expression to be rewritten as $\frac{1}{(2OP)^2} + \frac{1}{(2OR)^2} = \frac{1}{4} \left( \frac{1}{OP^2} + \frac{1}{OR^2} \right)$.

Given $P = \left( \frac{2\sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}} \right)$, the squared distance from the origin is $OP^2 = \left( \frac{2\sqrt{3}}{\sqrt{7}} \right)^2 + \left( \frac{6}{\sqrt{7}} \right)^2 = \frac{12}{7} + \frac{36}{7} = \frac{48}{7}$. The slope of $OP$ is calculated as $\frac{6/\sqrt{7}}{2\sqrt{3}/\sqrt{7}} = \sqrt{3}$. Since $PQ$ and $RS$ are mutually perpendicular, the slope of $OR$ must be the negative reciprocal, $-\frac{1}{\sqrt{3}}$. For any point $R(x, y)$ on the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$, substituting $y = -\frac{1}{\sqrt{3}}x$ results in $\frac{x^2}{4} + \frac{x^2}{27} = 1$. Solving this gives $x^2 \left( \frac{31}{108} \right) = 1$, so $x^2 = \frac{108}{31}$. The squared distance $OR^2 = x^2 + y^2 = x^2 + \frac{x^2}{3} = \frac{4}{3}x^2 = \frac{4}{3} \cdot \frac{108}{31} = \frac{144}{31}$.

Substituting these values into the expression yields $\frac{1}{4} \left( \frac{7}{48} + \frac{31}{144} \right) = \frac{1}{4} \left( \frac{21}{144} + \frac{31}{144} \right) = \frac{1}{4} \left( \frac{52}{144} \right) = \frac{13}{144}$. With $p=13$ and $q=144$, the sum $p+q = 157$.

The ellipse defined by the equation $\frac{x^2}{2} + \frac{y^2}{4} = 1$ is vertically oriented with semi-axes $a = \sqrt{2}$ and $b = 2$, which places its foci at the points $(0, \pm \sqrt{b^2 - a^2}) = (0, \pm \sqrt{2})$. Since $S$ is the focus on the negative major axis, its coordinates are $(0, -\sqrt{2})$. Any tangent line to this ellipse satisfies the equation $y = mx \pm \sqrt{2m^2 + 4}$. By substituting the coordinates of the intersection point $R(\sqrt{2}, 2\sqrt{2}-2)$ into this tangent equation, we determine the slopes of the lines passing through $R$, which results \in a vertical line $x = \sqrt{2}$ and a slanted line $y = -\sqrt{2}x + 2\sqrt{2}$.

The point of tangency $P$ for the vertical line $x = \sqrt{2}$ is found by substituting this x-value into the ellipse equation, giving $P(\sqrt{2}, 0)$. For the slanted tangent, the point of contact $Q$ is calculated using the property that a line $y = mx + c$ touches the ellipse at $(\frac{-a^2m}{c}, \frac{b^2}{c})$, which yields the coordinates $Q(1, \sqrt{2})$. The squared distance from $S(0, -\sqrt{2})$ to $P(\sqrt{2}, 0)$ is $SP^2 = (\sqrt{2}-0)^2 + (0 - (-\sqrt{2}))^2 = 2 + 2 = 4$. Similarly, the squared distance from $S$ to $Q(1, \sqrt{2})$ is $SQ^2 = (1-0)^2 + (\sqrt{2} - (-\sqrt{2}))^2 = 1 + 8 = 9$. The sum $SP^2 + SQ^2$ is therefore $4 + 9 = 13$.

The area of a triangle inscribed in the ellipse defined by $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ is determined by the coordinates of its vertices, which can be expressed parametrically as $(a \cos \theta, 2 \sin \theta)$. By placing one vertex at $(-a, 0)$ and the opposite side as a vertical chord, the base length is $4 \sin \theta$ and the height is $a + a \cos \theta$. The area function $A(\theta) = \frac{1}{2} (4 \sin \theta) (a + a \cos \theta) = 2a \sin \theta (1 + \cos \theta)$ is maximized by identifying where its derivative, $2a (2 \cos^2 \theta + \cos \theta - 1)$, is equal to zero.

The maximum area occurs when $\cos \theta = \frac{1}{2}$, which implies a corresponding $\sin \theta = \frac{\sqrt{3}}{2}$. Substituting these values into the area formula gives $2a \left(\frac{\sqrt{3}}{2}\right) \left(1 + \frac{1}{2}\right) = 6 \sqrt{3}$, which simplifies to $\frac{3a\sqrt{3}}{2} = 6 \sqrt{3}$ and confirms that $a = 4$. Given the semi-minor axis $b = 2$ and semi-major axis $a = 4$, the eccentricity is calculated using $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{16}}$, which results in $\frac{\sqrt{3}}{2}$.

The points where the ellipse intersects the coordinate axes correspond to its vertices, allowing us to directly determine the lengths of the semi-axes from the intercepts of the given lines. Setting $y=0$ in the first equation $\frac{x}{7} + \frac{y}{2\sqrt{6}} = 1$ identifies the $x$-axis vertex at $x=7$, which implies $a=7$ and $a^2=49$, while setting $x=0$ in the second equation $\frac{x}{7} - \frac{y}{2\sqrt{6}} = 1$ identifies the $y$-axis vertex at $y=-2\sqrt{6}$, which implies $b^2 = (-2\sqrt{6})^2 = 24$. Substituting these values into the eccentricity formula $e = \sqrt{1 - \frac{b^2}{a^2}}$ results in $e = \sqrt{1 - \frac{24}{49}}$, which simplifies to $\sqrt{\frac{25}{49}}$ or $\frac{5}{7}$.