Ellipse – JEE Main MathematicsPractice Questions & PYQs

The geometry of an ellipse is governed by the relationship between its axes, its foci, and the properties of its tangent lines. Finding the coordinates of a specific point of tangency relies on the condition that the slope of the tangent must be perpendicular to a given line, while the area of the triangle formed by the foci and that point is calculated using the distance between the foci as the base and the vertical coordinate of the point as the height.

For the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$, we identify the parameters $a^{2} = 8$ and $b^{2} = 4$. Using the relation $c^{2} = a^{2} - b^{2}$, we find $c = 2$, which locates the foci at 2 and -2 along the x-axis, establishing a total base length of 4 for the triangle. Since the tangent at P is perpendicular to the line $x+2y=0$, which has a slope of $-1/2$, the tangent must have a slope of 2. Applying this slope condition to the general tangent equation $\frac{x x_{1}}{8}+\frac{y y_{1}}{4}=1$ leads to the relationship $x_{1} = -4y_{1}$. Substituting this into the original ellipse equation shows that for a point in the second quadrant, P is located at $(-\frac{8}{3}, \frac{2}{3})$.

The area A of the triangle is calculated using the base of 4 and the height of $2/3$, resulting in $A = \frac{1}{2} \times 4 \times \frac{2}{3} = \frac{4}{3}$. With the eccentricity squared determined as $e^{2} = \frac{c^{2}}{a^{2}} = \frac{4}{8} = \frac{1}{2}$, we compute the final expression $(5-e^{2})A$ as $(5 - \frac{1}{2}) \times \frac{4}{3}$. This simplifies to $\frac{9}{2} \times \frac{4}{3}$, which results in 6.

The ellipse $x^2 + 4y^2 = 4$ simplifies to $\frac{x^2}{4} + y^2 = 1$, identifying its major axis extremities at $x = -2$ and $x = 2$, where the tangents are vertical lines. An arbitrary tangent to the ellipse at the point $(2 \cos \theta, \sin \theta)$ is given by $x \cos \theta + 2y \sin \theta = 2$, which intersects the vertical tangents at $B(-2, \cot \frac{\theta}{2})$ and $C(2, \tan \frac{\theta}{2})$. Using the diameter form of a circle, the equation is:
$(x+2)(x-2) + (y - \cot \frac{\theta}{2})(y - \tan \frac{\theta}{2}) = 0$
Expanding this expression leads to $x^2 + y^2 - y(\cot \frac{\theta}{2} + \tan \frac{\theta}{2}) - 3 = 0$. Substituting the coordinates $(\sqrt{3}, 0)$ into this final equation yields $3 + 0 - 0 - 3 = 0$, verifying that the circle passes through this location regardless of the parameter $\theta$.

The eccentricity $e$ of the ellipse $E_{1}$ satisfies the relation $e^{2} = 1 - \frac{b^{2}}{a^{2}}$. For the ellipse $E_{2}$, which has a semi-minor axis of length $a$ and foci at a distance $b$ from the center, the eccentricity $e$ must satisfy $e^{2} = 1 - \frac{a^{2}}{c^{2}}$, where $c$ represents the semi-major axis of $E_{2}$. Equating these two expressions for $e^{2}$ leads to $\frac{b^{2}}{a^{2}} = \frac{a^{2}}{c^{2}}$, which simplifies to $c = \frac{a^{2}}{b}$. Utilizing the focal distance relationship $b = ce$ for $E_{2}$, we substitute $c$ to obtain $\frac{b}{e} = \frac{a^{2}}{b}$, which simplifies to $e = \frac{b^{2}}{a^{2}}$. By substituting this expression for $e$ back into the first equation, we obtain $e = 1 - e^{2}$, resulting in the quadratic equation $e^{2} + e - 1 = 0$. Solving this equation for a positive value of $e$ yields $e = \frac{-1 + \sqrt{5}}{2}$.

The most effective way to determine the locus of mid-points for segments connected to an ellipse is to represent the ellipse using parametric coordinates. Any arbitrary point on the ellipse defined by $\frac{x^2}{4} + \frac{y^2}{9} = 1$ can be expressed as $(2\cos\theta, 3\sin\theta)$. By letting $(x, y)$ represent the midpoint of the line segment joining $(-3, -5)$ and this parametric point, we apply the midpoint formula to get $x = \frac{2\cos\theta - 3}{2}$ and $y = \frac{3\sin\theta - 5}{2}$.

Rearranging these equations to isolate the trigonometric terms gives $\cos\theta = \frac{2x + 3}{2}$ and $\sin\theta = \frac{2y + 5}{3}$. Since $\cos^2\theta + \sin^2\theta = 1$, we substitute these expressions to obtain $\frac{(2x + 3)^2}{4} + \frac{(2y + 5)^2}{9} = 1$. Clearing the fractions by multiplying the entire equation by $36$ results in $9(4x^2 + 12x + 9) + 4(4y^2 + 20y + 25) = 36$. Expanding these brackets yields $36x^2 + 108x + 81 + 16y^2 + 80y + 100 = 36$, which simplifies to $36x^2 + 16y^2 + 108x + 80y + 145 = 0$.

The equation of the tangent line to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at the specified point $(3 \sqrt{3} \cos \theta, \sin \theta)$ is represented by $\frac{x\cos\theta}{3\sqrt{3}} + y\sin\theta = 1$. By analyzing the axes intercepts of this line, we find the x-intercept to be $3\sqrt{3}\sec\theta$ and the y-intercept to be $\csc\theta$. The sum of these intercepts is therefore expressed by the function $f(\theta) = 3\sqrt{3}\sec\theta + \csc\theta$.

To determine the minimum value, differentiate $f(\theta)$ with respect to $\theta$ and equate the derivative to zero, resulting in $3\sqrt{3}\sec\theta\tan\theta - \csc\theta\cot\theta = 0$. This relationship simplifies to the equation $\frac{3\sqrt{3}\sin\theta}{\cos^{2}\theta} = \frac{\cos\theta}{\sin^{2}\theta}$, which rearranges to $\tan^{3}\theta = \frac{1}{3\sqrt{3}}$. Solving for $\theta$ within the given domain yields $\tan\theta = \frac{1}{\sqrt{3}}$, confirming that the minimum occurs at $\theta = \frac{\pi}{6}$.

The intersection points of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ and the circle $x^2 + y^2 = 4b$ must also lie on the curve $y^2 = 3x^2$, meaning the coordinates of these points satisfy all three equations simultaneously. Using the circle equation and the curve condition, we can substitute $x^2 = \frac{y^2}{3}$ into $x^2 + y^2 = 4b$ to find $\frac{y^2}{3} + y^2 = 4b$, which simplifies to $\frac{4}{3}y^2 = 4b$, giving $y^2 = 3b$ and consequently $x^2 = b$.

Substituting these values of $x^2$ and $y^2$ into the ellipse equation results in the expression $\frac{b}{16} + \frac{3b}{b^2} = 1$, which further simplifies to $\frac{b}{16} + \frac{3}{b} = 1$. Multiplying through by $16b$ yields the quadratic equation $b^2 + 48 = 16b$, or $b^2 - 16b + 48 = 0$. Factoring this equation gives $(b - 12)(b - 4) = 0$, and since the problem stipulates that $b > 4$, the only possible value is $b = 12$.

The intersection points of an ellipse and a circle are determined by solving their equations simultaneously, which requires first identifying the specific parameters of the ellipse from its given point and eccentricity. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the eccentricity condition $e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{3}$ implies $b^2 = \frac{2}{3}a^2$. Substituting this relationship and the coordinate point $(\sqrt{\frac{3}{2}}, 1)$ into the ellipse equation, $\frac{3}{2a^2} + \frac{1}{b^2} = 1$, allows us to solve for $a^2 = 3$ and $b^2 = 2$, establishing the ellipse as $\frac{x^2}{3} + \frac{y^2}{2} = 1$.

With the ellipse equation defined, the focus $(\alpha, 0)$ is found at $\alpha = \sqrt{a^2 - b^2} = 1$. The circle, centered at $(1, 0)$ with radius $\frac{2}{\sqrt{3}}$, has the equation $(x-1)^2 + y^2 = \frac{4}{3}$. Substituting $y^2 = 2(1 - \frac{x^2}{3})$ from the ellipse into the circle equation yields $(x-1)^2 + 2 - \frac{2x^2}{3} = \frac{4}{3}$, which simplifies to $x^2 - 6x + 5 = 0$. Solving this quadratic gives $x=1$ as the only valid value that maintains points on the ellipse, resulting in $y = \pm \frac{2}{\sqrt{3}}$. Consequently, the square of the distance between the two points $P$ and $Q$ is calculated as $(\frac{4}{\sqrt{3}})^2 = \frac{16}{3}$.

The pair of lines connecting the origin to the intersection points $P$ and $Q$ can be determined by homogenizing the curve equation $x^{2}+2y^{2}=2$ with the linear equation $x+y=1$. Setting the curve equation equal to $2(x+y)^{2}$ allows us to incorporate the line into the quadratic form, resulting in $x^{2}+2y^{2}=2(x^{2}+2xy+y^{2})$. Expanding this expression leads to $x^{2}+2y^{2}=2x^{2}+4xy+2y^{2}$, which simplifies to $x^{2}+4xy=0$ after canceling terms.

Factoring this homogeneous equation yields $x(x+4y)=0$, which identifies the two lines passing through the origin as the vertical axis $x=0$ and the line $y=-\frac{1}{4}x$. Because the line $x=0$ corresponds to the y-axis and the line $y=-\frac{1}{4}x$ lies at an angle of $\tan^{-1}\left(\frac{1}{4}\right)$ from the negative x-axis, the total angle subtended by the chord $PQ$ at the origin is $\frac{\pi}{2}+\tan^{-1}\left(\frac{1}{4}\right)$.

The position of an ellipse's directrices is intrinsically linked to the geometry of reflected light rays and the ellipse's eccentricity. To identify these lines, we first determine the equation of the reflected light ray which acts as the first directrix. Since the ray passes through $(2,1)$ and reflects off the $y$-axis to pass through $(5,3)$, we locate the image of $(2,1)$ across the $y$-axis at $(-2,1)$. Connecting $(-2,1)$ and $(5,3)$ gives a slope of $\frac{3-1}{5-(-2)} = \frac{2}{7}$, leading to the linear equation $y-1 = \frac{2}{7}(x+2)$, which simplifies to $2x-7y+11=0$.

For an ellipse with eccentricity $e=\frac{1}{3}$, the distance between the nearer focus and its corresponding directrix is given by the expression $\frac{a}{e}-ae$. Substituting the given distance of $\frac{8}{\sqrt{53}}$ into the expression $3a-\frac{a}{3} = \frac{8a}{3}$ allows us to solve for $a$, resulting in $a=\frac{3}{\sqrt{53}}$. The distance between the two parallel directrices of an ellipse is defined as $\frac{2a}{e}$, or $6a$, which equals $\frac{18}{\sqrt{53}}$.

Given that the second directrix must be parallel to the first, we express it as $2x-7y+\lambda=0$. The perpendicular distance between these two parallel lines is calculated using the formula $\frac{|\lambda-11|}{\sqrt{2^2+(-7)^2}}$, which simplifies to $\frac{|\lambda-11|}{\sqrt{53}}$. Equating this to the total distance between directrices, $\frac{18}{\sqrt{53}}$, we solve $|\lambda-11|=18$, resulting in $\lambda=29$ and $\lambda=-7$. Thus, the two possible equations for the other directrix are $2x-7y+29=0$ or $2x-7y-7=0$.

The tangency of a line to an ellipse is characterized by the condition that the quadratic equation formed by substituting the line's equation into the ellipse's equation has exactly one real root, meaning its discriminant must be zero. For an ellipse with axes parallel to the coordinate axes, centered at $(3, -4)$, the vertex at $(5, -4)$ gives the semi-major axis $a = 5 - 3 = 2$, and the focus at $(4, -4)$ indicates the product $ae = 4 - 3 = 1$, leading to an eccentricity $e = 1/2$. Consequently, the semi-minor axis squared is $b^2 = a^2(1 - e^2) = 4(1 - 1/4) = 3$, yielding the ellipse equation $\frac{(x-3)^2}{4} + \frac{(y+4)^2}{3} = 1$.

Substituting the tangent equation $y = mx - 4$ into this ellipse formula simplifies the vertical term, as $(y+4)$ becomes $mx - 4 + 4$, or simply $mx$. This transforms the equation into $\frac{(x-3)^2}{4} + \frac{(mx)^2}{3} = 1$, which expands to $3(x^2 - 6x + 9) + 4m^2x^2 = 12$. Rearranging this into a standard quadratic form gives:
$(3 + 4m^2)x^2 - 18x + 15 = 0$
For the line to be tangent, the discriminant of this quadratic must be zero, so $(-18)^2 - 4(3 + 4m^2)(15) = 0$. Simplifying this equation leads to $324 - 60(3 + 4m^2) = 0$, which simplifies to $324 = 180 + 240m^2$, or $144 = 240m^2$. Dividing both sides by 48 reveals that $3 = 5m^2$.

The intersection of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and the circle $x^{2}+y^{2}=3$ occurs at the point $(\frac{3}{2}, \frac{\sqrt{3}}{2})$ in the first quadrant. Since the angle between two curves is defined by the angle between their tangent lines at the point of intersection, we use implicit differentiation to determine the slopes of these tangents. For the ellipse, the slope is $m_{1} = -\frac{x}{9y}$, and for the circle, the slope is $m_{2} = -\frac{x}{y}$.

Evaluating these expressions at $(\frac{3}{2}, \frac{\sqrt{3}}{2})$ gives $m_{1} = -\frac{1}{3\sqrt{3}}$ and $m_{2} = -\sqrt{3}$. Substituting these into the angle formula $\tan \theta = \left|\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}\right|$, the numerator becomes $\frac{8}{3\sqrt{3}}$ and the denominator becomes $\frac{4}{3}$. This simplifies to $\tan \theta = \frac{2}{\sqrt{3}}$.

The geometry of a tangent line to an ellipse is most efficiently handled by expressing a general point on the curve $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{4a^{2}}=1$ using parametric coordinates $(b \cos \theta, 2a \sin \theta)$. By deriving the equation of the tangent at this specific point, $\frac{x \cos \theta}{b} + \frac{y \sin \theta}{2a} = 1$, we can readily find the intercepts where the line meets the coordinate axes. Setting the variables to zero in turn reveals the intercepts at $x = b \sec \theta$ and $y = 2a \csc \theta$.

These two intercepts serve as the base and height of the right triangle formed with the origin, resulting in an area calculation of $\frac{1}{2}(b \sec \theta)(2a \csc \theta)$. Through trigonometric identities, this simplifies to $\frac{ab}{\sin \theta \cos \theta}$, which is equivalent to $\frac{2ab}{\sin 2\theta}$. Since the sine function is bounded by a maximum value of 1, the area of the triangle is minimized precisely when $\sin 2\theta$ reaches its peak. Consequently, the minimum area is $2ab$, which identifies the constant $k$ as 2.

Any point on the ellipse $2x^2 + y^2 = 1$ can be represented parametrically as $x = \frac{1}{\sqrt{2}}\cos\theta$ and $y = \sin\theta$. The equation of the normal line to the ellipse at such a point is given by
$\frac{x\sec\theta}{\sqrt{2}} - y\csc\theta = -\frac{1}{2}$
By rearranging this into the intercept form $\frac{x}{X_0} + \frac{y}{Y_0} = 1$, we identify the x-intercept as $-\frac{\cos\theta}{\sqrt{2}}$ and the y-intercept as $\frac{\sin\theta}{2}$. Setting the x-intercept equal to the given value $-\frac{1}{3\sqrt{2}}$ yields $\cos\theta = \frac{1}{3}$. Using the identity $\sin^2\theta = 1 - \cos^2\theta$, we find $\sin\theta = \frac{2\sqrt{2}}{3}$. Since the y-intercept corresponds to $\beta = \frac{\sin\theta}{2}$, substituting the value of $\sin\theta$ leads to $\beta = \frac{\sqrt{2}}{3}$.

The geometry of an ellipse defined by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ dictates that when the minor axis lies along the y-axis, the vertical semi-axis length $b$ satisfies $2b = \frac{4}{\sqrt{3}}$, resulting in $b = \frac{2}{\sqrt{3}}$ and $b^2 = \frac{4}{3}$. To find the major axis length $a$, we examine the tangency condition for the line $x + 6y = 8$, which rearranges to the slope-intercept form $y = -\frac{1}{6}x + \frac{4}{3}$.

For any line to be tangent to an ellipse, it must satisfy the condition $c^2 = a^2 m^2 + b^2$. Substituting the values $m = -\frac{1}{6}$, $c = \frac{4}{3}$, and $b^2 = \frac{4}{3}$ yields the following equation:
$\left(\frac{4}{3}\right)^2 = a^2\left(-\frac{1}{6}\right)^2 + \frac{4}{3}$
Solving this equation results in $\frac{16}{9} = \frac{a^2}{36} + \frac{12}{9}$, which simplifies to $\frac{a^2}{36} = \frac{4}{9}$, establishing $a^2 = 16$. With $a^2 = 16$ and $b^2 = \frac{4}{3}$ now known, the eccentricity is calculated using $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting these values leads to $e = \sqrt{1 - \frac{4/3}{16}}$, which simplifies to $\sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}}$, ultimately resulting in $\frac{1}{2} \sqrt{\frac{11}{3}}$.

The relationship between the dimensions of an ellipse and its eccentricity is rooted in the definition of the latus rectum, $\frac{2b^2}{a} = 10$. This equality immediately implies that $b^2 = 5a$. Separately, the eccentricity is determined by finding the vertex of the function $\phi(t) = \frac{5}{12} + t - t^2$. By rewriting this as $\frac{8}{12} - (t - \frac{1}{2})^2$, we identify the maximum value as $\frac{2}{3}$, meaning the eccentricity is $e = \frac{2}{3}$.

Substituting $e^2 = \frac{4}{9}$ into the eccentricity formula $\frac{b^2}{a^2} = 1 - e^2$ results in $\frac{5a}{a^2} = 1 - \frac{4}{9}$. Simplifying this expression leads to $\frac{5}{a} = \frac{5}{9}$, which means $a = 9$ and subsequently $a^2 = 81$. Using the earlier relation $b^2 = 5a$, we find $b^2 = 45$. Adding these values together yields $a^2 + b^2 = 81 + 45 = 126$.

The locus of the foot of the perpendicular drawn from either focus to any tangent of an ellipse is consistently its auxiliary circle, defined by the equation $x^{2}+y^{2}=a^{2}$. For the given ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$, the semi-major axis squared is $a^{2}=4$, which leads directly to the auxiliary circle equation $x^{2}+y^{2}=4$. By testing the provided options to determine which point satisfies this relationship, we find that for the point $(-1,\sqrt{3})$, the calculation $(-1)^{2}+(\sqrt{3})^{2}=1+3=4$ holds true. Consequently, this coordinate is the only one among the choices that lies on the locus.

The geometry of an ellipse is governed by its eccentricity $e$ and semi-major axis $a$, where the distance between its foci is expressed as $2ae$ and the distance between its directrices is $\frac{2a}{e}$. Given these distances are 6 and 12 respectively, we establish the equations $ae = 3$ and $\frac{a}{e} = 6$. Multiplying these two equations provides $a^2 = 18$, and dividing the second by the first yields $e^2 = \frac{1}{2}$. We then calculate the square of the semi-minor axis using $b^2 = a^2(1 - e^2)$, which simplifies to $b^2 = 18(1 - \frac{1}{2}) = 9$. Using the formula $\frac{2b^2}{a}$ for the length of the latus rectum, we compute $\frac{2 \times 9}{3\sqrt{2}}$, which reduces to $3\sqrt{2}$.

The defining characteristic of an ellipse is that for any point on the curve, the sum of the distances to the two fixed focal points remains constant and equal to the length of the major axis. Transforming the equation $9x^2 + 16y^2 = 144$ by dividing by 144 yields the standard form $\frac{x^2}{16} + \frac{y^2}{9} = 1$, where $a^2 = 16$ and $b^2 = 9$. Calculating the focal distance $c = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}$ confirms that the points $A(\sqrt{7}, 0)$ and $B(-\sqrt{7}, 0)$ are exactly the foci of the ellipse. Therefore, the sum $PA + PB$ corresponds to the length of the major axis, $2a$, which results in $2 \times 4 = 8$.

The geometry of an ellipse defines the normal line at any point using the eccentricity and the axis lengths. When a line is perpendicular to the tangent at a specific coordinate on the latus rectum, its equation can be expressed in terms of the semi-major axis $a$ and eccentricity $e$. By substituting the coordinates of the endpoint of the minor axis, $(0, b)$, into the equation of this normal, we establish a direct algebraic constraint between the ellipse parameters.

The normal equation at the point $(ae, \frac{b^2}{a})$ on the ellipse is given by:
$\frac{a^2x}{ae} - \frac{b^2y}{b^2/a} = a^2e^2$
This expression simplifies to $\frac{ax}{e} - ay = a^2e^2$, which can be reduced to $\frac{x}{e} - y = ae^2$. Applying the condition that this line passes through the minor axis extremity $(0, b)$, we substitute these coordinates to arrive at $-b = ae^2$. Squaring both sides produces the relationship $b^2 = a^2e^4$. Substituting the standard identity $b^2 = a^2(1-e^2)$ leads to the equation $a^2(1-e^2) = a^2e^4$. Dividing by $a^2$ yields $1-e^2 = e^4$, which rearranges into the final relationship:
$e^4 + e^2 - 1 = 0$

The condition for a line $y = mx + c$ to be tangent to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is expressed by the relation $c^2 = a^2m^2 + b^2$. Transforming the equation of the line $3x + 4y = 12\sqrt{2}$ into slope-intercept form, we get $y = -\frac{3}{4}x + 3\sqrt{2}$, which identifies the slope $m = -\frac{3}{4}$ and the intercept $c = 3\sqrt{2}$.

Substituting $c^2 = 18$, $m^2 = \frac{9}{16}$, and $b^2 = 9$ into the tangency condition, we obtain $18 = a^2(\frac{9}{16}) + 9$. Subtracting $9$ from both sides results in $9 = \frac{9}{16}a^2$, which simplifies to $a^2 = 16$, meaning $a = 4$.

With the values of $a$ and $b$ identified, the eccentricity $e$ is determined using the relation $b^2 = a^2(1 - e^2)$, which becomes $9 = 16(1 - e^2)$, leading to $1 - e^2 = \frac{9}{16}$ or $e = \frac{\sqrt{7}}{4}$. The distance between the foci of the ellipse is defined as $2ae$, so substituting our values yields $2 \times 4 \times \frac{\sqrt{7}}{4}$, resulting in $2\sqrt{7}$.