Ellipse – JEE Main MathematicsPractice Questions & PYQs

An ellipse centered at the origin with directrix $x = 4$ and eccentricity $e = \frac{1}{2}$ relates to its semi-major axis $a$ through the formula $x = \frac{a}{e}$, which yields $a = 2$ and $a^2 = 4$. Utilizing the relationship $b^2 = a^2(1 - e^2)$, we find $b^2 = 4(1 - \frac{1}{4}) = 3$, constructing the ellipse equation as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Substituting $x = 1$ into this equation reveals $\frac{1}{4} + \frac{\beta^2}{3} = 1$, which gives $\beta^2 = \frac{9}{4}$, confirming that $\beta = \frac{3}{2}$ given the condition $\beta > 0$ and placing point $P$ at $(1, \frac{3}{2})$.

The equation of the normal at point $(x_1, y_1)$ for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$. Substituting the values $a^2 = 4$, $b^2 = 3$, $x_1 = 1$, and $y_1 = \frac{3}{2}$ leads to $\frac{4x}{1} - \frac{3y}{3/2} = 4 - 3$, which simplifies to $4x - 2y = 1$.

For an ellipse centered at the origin with its major axis along the x-axis, the length of the latus rectum is given by the relation $\frac{2b^2}{a} = 8$, which simplifies to $b^2 = 4a$. The problem specifies that the distance between the foci, $2ae$, is equal to the length of the minor axis, $2b$, implying $ae = b$ or $e = \frac{b}{a}$. Substituting this relationship into the standard definition of eccentricity, $e^2 = 1 - \frac{b^2}{a^2}$, yields $e^2 = 1 - e^2$, or $2e^2 = 1$, which confirms the eccentricity is $e = \frac{1}{\sqrt{2}}$.

Since $e^2 = \frac{b^2}{a^2} = \frac{1}{2}$, it follows that $b^2 = \frac{a^2}{2}$. Substituting the earlier result $b^2 = 4a$ into this equation gives $\frac{a^2}{2} = 4a$, which leads to $a = 8$ and subsequently $b^2 = 32$. The equation of the ellipse is therefore:

$\frac{x^2}{64} + \frac{y^2}{32} = 1$

Testing the coordinate pair $(4\sqrt{3}, 2\sqrt{2})$ in this equation confirms that $\frac{(4\sqrt{3})^2}{64} + \frac{(2\sqrt{2})^2}{32} = \frac{48}{64} + \frac{8}{32} = \frac{3}{4} + \frac{1}{4} = 1$, meaning this point lies on the ellipse.

In an ellipse with foci at $S$ and $S'$ and semi-minor axis $b$, the right-angled triangle condition at $B$ means the slopes of $SB$ and $S'B$ are negative reciprocals, which simplifies to the geometric property $b^2 = a^2e^2$. Since the coordinates of the foci are $(\pm ae, 0)$ and the vertex $B$ is $(0, b)$, the distance $SB$ is given by $\sqrt{a^2e^2 + b^2}$. The area of the right-angled triangle $\Delta S'BS$ is $\frac{1}{2} \times SB \times S'B$, and since the triangle is isosceles with $SB = S'B$, the area is $\frac{1}{2} SB^2$. Given the area is $8$, we have $SB^2 = 16$, which implies $a^2e^2 + b^2 = 16$.

Substituting the geometric property $a^2e^2 = b^2$ into this equation yields $2b^2 = 16$, or $b^2 = 8$, which means $b = 2\sqrt{2}$. Using the fundamental ellipse relation $b^2 = a^2(1 - e^2)$, or $b^2 = a^2 - a^2e^2$, we substitute the known values $b^2 = 8$ and $a^2e^2 = 8$ to find $8 = a^2 - 8$, resulting in $a^2 = 16$ and $a = 4$. The length of the latus rectum, defined by the formula $\frac{2b^2}{a}$, is calculated as:

$\frac{2b^2}{a} = \frac{2(8)}{4} = 4$

The ellipse equation $x^2 + 2y^2 = 2$ is equivalent to the standard form $\frac{x^2}{2} + y^2 = 1$, identifying the parameters $a = \sqrt{2}$ and $b = 1$. A tangent line at any point defined by angle $\theta$ follows the equation $\frac{x \cos \theta}{\sqrt{2}} + y \sin \theta = 1$, which intercepts the coordinate axes at the points $(\sqrt{2} \sec \theta, 0)$ and $(0, \csc \theta)$. The midpoint $(x, y)$ of this intercepted segment is calculated as $x = \frac{\sqrt{2} \sec \theta}{2}$ and $y = \frac{\csc \theta}{2}$, which can be rearranged to $\cos \theta = \frac{1}{\sqrt{2} x}$ and $\sin \theta = \frac{1}{2 y}$. Substituting these trigonometric terms into the identity $\cos^2 \theta + \sin^2 \theta = 1$ yields the equation of the required curve:

$\frac{1}{2x^2} + \frac{1}{4y^2} = 1$

The slope of the normal to the ellipse $3x^2 + 4y^2 = 12$ identifies the location of point $P$. By rewriting the equation in the standard form $\frac{x^2}{4} + \frac{y^2}{3} = 1$, the parameters are $a=2$ and $b=\sqrt{3}$. The slope of the normal at any point $P(2\cos\theta, \sqrt{3}\sin\theta)$ is given by the expression $\frac{a}{b}\tan\theta$, which must be parallel to the line $2x + y = 4$, implying a slope of $-2$. Setting $\frac{2}{\sqrt{3}}\tan\theta = -2$ yields $\tan\theta = -\sqrt{3}$, and since the point lies on the ellipse, this leads to $\cos\theta = -\frac{1}{2}$ and $\sin\theta = \frac{\sqrt{3}}{2}$, fixing $P$ at $(-1, \frac{3}{2})$. Calculating the length $PQ$ between $P(-1, \frac{3}{2})$ and $Q(4, 4)$ using the distance formula $\sqrt{(4 - (-1))^2 + (4 - \frac{3}{2})^2}$ results in $\sqrt{5^2 + (\frac{5}{2})^2}$, which simplifies to $\sqrt{\frac{125}{4}}$ or $\frac{5\sqrt{5}}{2}$.

The line tangent to the parabola $y^{2}=x$ at a point $(x_{0}, y_{0})$ can be expressed in the slope-intercept form $y=mx+c$. Given the parabola equation $y^{2}=4ax$ with $a=\frac{1}{4}$, any tangent with slope $m$ satisfies the equation $y=mx+\frac{1}{4m}$, identifying the intercept as $c=\frac{1}{4m}$. For this line to also be tangent to the ellipse $x^{2}+2y^{2}=1$, it must satisfy the standard tangency condition $c^{2}=a^{2}m^{2}+b^{2}$. By substituting the ellipse parameters $a^{2}=1$ and $b^{2}=\frac{1}{2}$ into this condition, we obtain $c^{2}=m^{2}+\frac{1}{2}$.

Equating these two expressions for $c^{2}$ leads to the equation $\frac{1}{16m^{2}}=m^{2}+\frac{1}{2}$, which simplifies to the biquadratic form $16m^{4}+8m^{2}-1=0$. Solving for $m^{2}$ using the quadratic formula provides $m^{2}=\frac{-8+\sqrt{64+64}}{32}$, which simplifies to $m^{2}=\frac{\sqrt{2}-1}{4}$. The $x$-coordinate of the point of tangency on the parabola is defined by $x_{0}=\frac{1}{4m^{2}}$. Substituting the calculated value of $m^{2}$ into this expression results in $x_{0}=\frac{1}{\sqrt{2}-1}$, which simplifies to $\sqrt{2}+1$. Since the problem identifies this coordinate as $\alpha$, the value is $\sqrt{2}+1$.

Since the focus is located on the y-axis at $(0, 5\sqrt{3})$, we are dealing with a vertical ellipse where the semi-major axis $b$ and semi-minor axis $a$ define the geometry. The difference between the major axis length $2b$ and the minor axis length $2a$ is 10, which simplifies to $b - a = 5$. Given the focus coordinate $be$ satisfies the relation $b^2e^2 = b^2 - a^2$, and knowing the focus is at $5\sqrt{3}$, we establish that $b^2 - a^2 = (5\sqrt{3})^2 = 75$. Factoring the difference of squares gives $(b - a)(b + a) = 75$, and substituting $b - a = 5$ leads to $b + a = 15$. Solving these simultaneous linear equations, $b - a = 5$ and $b + a = 15$, yields $b = 10$ and $a = 5$. The length of the latus rectum for a vertical ellipse is determined by the formula $\frac{2a^2}{b}$, which evaluates to $\frac{2(5)^2}{10} = \frac{50}{10} = 5$.

The classification of the conic section defined by $\frac{y^2}{1+r} - \frac{x^2}{1-r} = 1$ depends on the condition $r > 1$, which makes the term $1-r$ negative. By rewriting the equation as $\frac{y^2}{1+r} + \frac{x^2}{r-1} = 1$, we see that it takes the standard form of an ellipse with semi-axis squares $b^2 = 1+r$ and $a^2 = r-1$. The eccentricity $e$ is calculated using the formula $e = \sqrt{1 - \frac{a^2}{b^2}}$, which yields $e = \sqrt{1 - \frac{r-1}{r+1}} = \sqrt{\frac{r+1-(r-1)}{r+1}} = \sqrt{\frac{2}{r+1}}$.

Two lines are perpendicular if the product of their slopes is $-1$. For the ellipse $4x^2 + y^2 = 8$, the equation of the tangent at any point $(x_1, y_1)$ is given by $4x x_1 + y y_1 = 8$. Applying this to the point $(1, 2)$, we obtain $4x + 2y = 8$, which simplifies to $y = -2x + 4$, indicating a slope of $-2$. For the tangent at $(a, b)$ to be perpendicular, its slope must be the negative reciprocal, $\frac{1}{2}$. The general tangent equation at $(a, b)$ is $4ax + by = 8$, so the slope is $-\frac{4a}{b}$. Setting $-\frac{4a}{b} = \frac{1}{2}$ results in the relationship $b = -8a$. Since the point $(a, b)$ must satisfy the ellipse equation $4a^2 + b^2 = 8$, substituting $b = -8a$ gives $4a^2 + (-8a)^2 = 8$, which simplifies to $68a^2 = 8$, and leads to $a^2 = \frac{2}{17}$.

The equations for the tangent and normal lines at point $P(2,2)$ on the ellipse $3x^2 + 5y^2 = 32$ are derived from the slope of the curve, which is obtained via implicit differentiation as $6x + 10y \frac{dy}{dx} = 0$. Substituting the coordinates of $P$, the derivative becomes $\frac{dy}{dx} = -\frac{3(2)}{5(2)} = -\frac{3}{5}$, defining the slope of the tangent line. Consequently, the tangent equation is $y - 2 = -\frac{3}{5}(x - 2)$, which simplifies to $3x + 5y = 16$, while the normal line, possessing a perpendicular slope of $\frac{5}{3}$, follows the equation $y - 2 = \frac{5}{3}(x - 2)$, or $5x - 3y = 4$.

The x-intercepts $Q$ and $R$ are identified by setting $y=0$ in these linear equations, resulting in $x = \frac{16}{3}$ for $Q$ and $x = \frac{4}{5}$ for $R$. With these points located on the x-axis, the base of the triangle $QR$ measures $\frac{16}{3} - \frac{4}{5} = \frac{68}{15}$. Given that the height of the triangle is the $y$-coordinate of point $P$, which is $2$, the area is calculated as $\frac{1}{2} \times \frac{68}{15} \times 2$, yielding $\frac{68}{15}$.

The standard equation of an ellipse centered at the origin with its foci on the vertical axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $b > a$. The foci at $(0, 2)$ and $(0, -2)$ identify the distance from the center to each focus as $c = 2$, and the minor axis length of 4 corresponds to $2a = 4$, which means $a = 2$ and $a^2 = 4$.

Using the relationship $c^2 = b^2 - a^2$, we substitute the known values to find $2^2 = b^2 - 4$, simplifying to $b^2 = 8$. This yields the specific equation $\frac{x^2}{4} + \frac{y^2}{8} = 1$. Checking the point $(\sqrt{2}, 2)$ in this equation gives $\frac{(\sqrt{2})^2}{4} + \frac{2^2}{8} = \frac{2}{4} + \frac{4}{8} = 1$, which confirms the point lies on the ellipse.

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 defined by the relationship $c^2 = a^2m^2 + b^2$. For the given line $x - 2y = 12$, rewriting it in slope-intercept form gives $y = \frac{1}{2}x - 6$, where the slope $m = \frac{1}{2}$ and the $y$-intercept $c = -6$. Substituting these values into the tangency condition yields $36 = \frac{a^2}{4} + b^2$, providing a direct relationship between $a^2$ and $b^2$.

Because the point of tangency $(3, -\frac{9}{2})$ must lie on the ellipse, substituting these coordinates into the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ produces $\frac{9}{a^2} + \frac{81}{4b^2} = 1$. Solving these two simultaneous equations, $\frac{a^2}{4} + b^2 = 36$ and $\frac{9}{a^2} + \frac{81}{4b^2} = 1$, yields the values $b^2 = 27$ and $a^2 = 36$. Applying these values to the formula for the length of the latus rectum, $\frac{2b^2}{a}$, results in $\frac{2(27)}{\sqrt{36}}$, which simplifies to 9.

The length of the latus rectum of an ellipse is given by $\frac{2b^2}{a}$, while the distance between a focus and its nearest vertex is defined as $a(1-e)$, where $a$ denotes the semi-major axis, $b$ the semi-minor axis, and $e$ the eccentricity. Based on the given values, $\frac{2b^2}{a} = 4$, which simplifies to $b^2 = 2a$, and $a(1-e) = \frac{3}{2}$.

By utilizing the standard ellipse identity $b^2 = a^2(1-e^2)$, the substitution $b^2 = 2a$ results in $2a = a^2(1-e^2)$, which simplifies to $2 = a(1-e)(1+e)$. Substituting the known value $a(1-e) = \frac{3}{2}$ into this expression gives $2 = \frac{3}{2}(1+e)$. Consequently, $4 = 3(1+e)$, which leads to $4 = 3 + 3e$ or $3e = 1$, yielding the eccentricity $e = \frac{1}{3}$.

The geometric properties of an ellipse, specifically the relationship between its eccentricity and the position of its directrix, define its dimensions. Given the eccentricity $e = \frac{1}{2}$ and the directrix equation $x = -4$, we use the directrix formula $x = -\frac{a}{e} = -4$ to determine that $a = 2$. Applying the relationship $b^2 = a^2(1 - e^2)$, we calculate $b^2 = 4(1 - \frac{1}{4}) = 3$, which establishes the equation of the ellipse as $\frac{x^2}{4} + \frac{y^2}{3} = 1$.

Differentiating this equation implicitly with respect to $x$ provides the expression for the tangent slope, $\frac{2x}{4} + \frac{2y}{3} \frac{dy}{dx} = 0$, which simplifies to $\frac{dy}{dx} = -\frac{3x}{4y}$. At the specific point $\left(1, \frac{3}{2}\right)$, the tangent slope evaluates to $-\frac{3(1)}{4(3/2)} = -\frac{1}{2}$. Because the normal is perpendicular to the tangent at the point of contact, its slope is the negative reciprocal, $2$. Using the point-slope form $y - \frac{3}{2} = 2(x - 1)$, we arrive at the linear equation $2y - 3 = 4x - 4$, which simplifies to $4x - 2y = 1$.

The standard equation of an ellipse centered at the origin with axes along the coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Substituting the coordinates $(4, -1)$ and $(-2, 2)$ into this equation gives the system $\frac{16}{a^2} + \frac{1}{b^2} = 1$ and $\frac{4}{a^2} + \frac{4}{b^2} = 1$. By setting $X = \frac{1}{a^2}$ and $Y = \frac{1}{b^2}$, these equations become $16X + Y = 1$ and $4X + 4Y = 1$, where the latter simplifies to $X + Y = \frac{1}{4}$. Subtracting the simplified second equation from the first yields $15X = \frac{3}{4}$, which implies $X = \frac{1}{20}$, and substituting this value back leads to $Y = \frac{1}{5}$. With $a^2 = 20$ and $b^2 = 5$, the eccentricity is determined by the relation $b^2 = a^2(1 - e^2)$, resulting in $5 = 20(1 - e^2)$. This simplifies to $1 - e^2 = \frac{1}{4}$, which gives $e^2 = \frac{3}{4}$ and finally $e = \frac{\sqrt{3}}{2}$.

An ellipse centered at the origin with its major axis along the x-axis possesses vertices located at $(\pm a, 0)$ and $(0, \pm b)$, which form a rhombus when connected to define the inscribed quadrilateral, with an area calculated as $2ab$. Because the distance between the foci is $2ae = 6$ and the eccentricity is $e = \frac{3}{5}$, the semi-major axis is determined by solving $a(\frac{3}{5}) = 3$, yielding $a = 5$. The relation $b^2 = a^2(1 - e^2)$ then provides $b^2 = 25(1 - \frac{9}{25}) = 16$, resulting in $b = 4$. Substituting these values into the area formula $2ab$ produces $2(5)(4) = 40$ square units.

Any point on the ellipse $\frac{x^2}{27}+\frac{y^2}{3}=1$ can be expressed using parametric coordinates as $(3\sqrt{3}\cos\theta, \sqrt{3}\sin\theta)$. The tangent to the ellipse at this point is given by the equation $\frac{x\cos\theta}{3\sqrt{3}} + \frac{y\sin\theta}{\sqrt{3}} = 1$. By setting the $y$ and $x$ coordinates to zero respectively, we find the intercepts on the axes are $A = (3\sqrt{3}\sec\theta, 0)$ and $B = (0, \sqrt{3}\csc\theta)$.

The area of the right-angled triangle formed by the origin and these intercepts is calculated as $\frac{1}{2} \times OA \times OB$. Substituting the intercept values, the area becomes $\frac{1}{2} \cdot (3\sqrt{3}\sec\theta) \cdot (\sqrt{3}\csc\theta)$. Simplifying this product yields $\frac{9}{2\sin\theta\cos\theta}$, which can be rewritten using the double-angle identity as $\frac{9}{\sin2\theta}$. Since the maximum value of $\sin2\theta$ is $1$, the minimum area of the triangle occurs when the denominator is at its peak, resulting in a value of $9$.

The area of the quadrilateral formed by tangents at the endpoints of the latera recta relies on the symmetry of the ellipse, where calculating the area of the region in one quadrant and multiplying by four yields the total result. For the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$, the focus occurs at $x=ae=3 \times \frac{2}{3}=2$, which corresponds to a vertical endpoint on the ellipse at $(2, \frac{5}{3})$. The equation of the tangent at this point is $\frac{2x}{9} + \frac{y}{3} = 1$, which intercepts the axes at $\frac{9}{2}$ and $3$. This defines a right triangle in the first quadrant with an area of $\frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$, and multiplying this value by four provides a final area of 27.

The condition for a line $y = mx + c$ to be tangent to the ellipse $\frac{x^2}{6} + \frac{y^2}{2} = 1$ is $c^2 = 6m^2 + 2$. If the foot of the perpendicular from the center $(0, 0)$ to such a tangent is $(h, k)$, the line connecting the center to this point has a slope of $k/h$, implying the tangent line has a perpendicular slope of $m = -h/k$. The equation of the tangent line passing through $(h, k)$ is $y - k = -\frac{h}{k}(x - h)$, which simplifies to $y = -\frac{h}{k}x + \frac{h^2 + k^2}{k}$. Substituting $m = -h/k$ and the intercept $c = \frac{h^2 + k^2}{k}$ into the tangency condition results in $(\frac{h^2 + k^2}{k})^2 = 6(-\frac{h}{k})^2 + 2$. Clearing the denominators by multiplying by $k^2$ gives:
$(x^2 + y^2)^2 = 6x^2 + 2y^2$

The ellipse defined by $\frac{x^2}{16}+\frac{y^2}{9}=1$ has a major axis along the x-axis, so its foci are situated at $(\pm ae, 0)$ where $a^2e^2 = a^2 - b^2 = 16 - 9 = 7$, placing them at $(\pm \sqrt{7}, 0)$. Given the circle has its center at $(0, 3)$ and must pass through these foci, we determine the square of its radius by calculating the squared distance from the center to the point $(\sqrt{7}, 0)$, which yields $(\sqrt{7}-0)^2 + (0-3)^2 = 7 + 9 = 16$. Applying the standard circle equation form $(x-h)^2 + (y-k)^2 = r^2$ with center $(0, 3)$ and $r^2 = 16$ results in $x^2 + (y-3)^2 = 16$, which simplifies to $x^2+y^2-6y-7=0$.