Parabola – JEE Main MathematicsPractice Questions & PYQs

The parabola $y^2 = 12x$ has the standard form $y^2 = 4ax$ with $a=3$, allowing us to represent point $P$ as $(3t^2, 6t)$ and its projection $N$ on the axis as $(3t^2, 0)$. Because $M$ is the midpoint of $PN$, its coordinates are $(3t^2, 3t)$, and the horizontal line passing through $M$ is $y = 3t$. Solving for the intersection of this line with the parabola reveals point $Q$ at $(\frac{3}{4}t^2, 3t)$. The line $NQ$ passing through $(3t^2, 0)$ and $(\frac{3}{4}t^2, 3t)$ has a slope of $-\frac{4}{3t}$, leading to the equation $y = -\frac{4}{3t}(x - 3t^2)$. Setting $x=0$ shows the $y$-intercept is $4t$, which matches $\frac{4}{3}$ when $t=\frac{1}{3}$. Substituting this parameter into the distance $MQ = |3t^2 - \frac{3}{4}t^2| = \frac{9}{4}t^2$ results in $MQ = \frac{1}{4}$.

To determine the center of this circle, we first identify the tangent line to the parabola $y=x^2$ at the point $(2,4)$, where the derivative $\frac{dy}{dx}=2x$ gives a slope of $4$, resulting in the linear equation $4x - y - 4 = 0$. Any circle touching the parabola at this specific point belongs to the family defined by $(x-2)^2 + (y-4)^2 + \lambda(4x - y - 4) = 0$, and since the circle passes through $(0,1)$, substituting these coordinates gives $4 + 9 - 5\lambda = 0$, leading to $\lambda = \frac{13}{5}$. By expanding the family equation $x^2 + y^2 + x(4\lambda - 4) + y(-\lambda - 8) + (20 - 4\lambda) = 0$, the general coordinates for the center are derived from the coefficients of $x$ and $y$ as $\left(2 - 2\lambda, \frac{\lambda + 8}{2}\right)$. Substituting $\lambda = \frac{13}{5}$ into these center expressions produces the final coordinates $\left(2 - \frac{26}{5}, \frac{13/5 + 8}{2}\right)$, which simplifies to $\left(\frac{-16}{5}, \frac{53}{10}\right)$.

Finding the locus of a point involves establishing an algebraic relationship between its coordinates by eliminating any independent parameters using the geometric definitions provided. Since the point lies on the parabola $x^2 = 4y$, we can represent any position on it using the parametric form $(2t, t^2)$. Applying the section formula to divide the segment joining this point and $(0, -1)$ in a 1:2 ratio, we define the coordinates of the locus point $(x, y)$ as $x = \frac{1(2t) + 2(0)}{1+2} = \frac{2t}{3}$ and $y = \frac{1(t^2) + 2(-1)}{1+2} = \frac{t^2 - 2}{3}$. Isolating the parameter in the first equation gives $t = \frac{3x}{2}$, and substituting this into the second expression yields $3y = \left(\frac{3x}{2}\right)^2 - 2$. Multiplying the entire equation by 4 to clear the denominator results in $12y = 9x^2 - 8$, which rearranges to the final locus equation $9x^2 - 12y = 8$

Any point $P$ on the parabola $y^2 = x$ can be parameterized as $(t^2, t)$ for some $t \neq 0$. Because the line $y = mx$ intersects the parabola at $P$, substituting $P$ into the line equation gives $t = m(t^2)$, which establishes the relationship $m = 1/t$. The tangent line to the parabola at $P(t^2, t)$ is given by $2ty = x + t^2$, and setting $y = 0$ shows that this tangent intersects the x-axis at $Q(-t^2, 0)$. The area of $\Delta OPQ$ with vertices $O(0,0)$, $P(t^2, t)$, and $Q(-t^2, 0)$ is calculated as $\frac{1}{2} |(t^2)(0) - (-t^2)(t)| = \frac{1}{2} t^3$. Equating this area to 4 results in $t^3 = 8$, which means $t = 2$, and consequently, $m = 0.50$.

The parabola $y^2 = 8x$ follows the standard form $y^2 = 4ax$, which identifies the parameter $a = 2$. Since any point on the parabola can be represented in parametric form as $(at^2, 2at)$, we find the parameter $t_1$ for point $A(\frac{1}{2}, -2)$ by solving $2at_1 = -2$, resulting in $t_1 = -\frac{1}{2}$. For a focal chord, the parameters of the endpoints satisfy the property $t_1 t_2 = -1$, which implies $t_2 = 2$.

Substituting $t_2 = 2$ into the parametric coordinates $(at^2, 2at)$ gives the location of point $B$ as $(8, 8)$. The equation of the tangent at any point $(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$. Applying this formula with $a = 2$ and point $B(8, 8)$ yields $8y = 4(x + 8)$, which simplifies to the final equation $x - 2y + 8 = 0$.

The equation of a tangent to a parabola of the form $y^2=4a(x-h)$ with slope $m$ is given by $y=m(x-h) + a/m$. For the first parabola $y^2=4(x+1)$, the vertex is at $(-1, 0)$ and $4a=4$, which implies $a=1$. Consequently, any tangent to this curve can be expressed as $y=m(x+1)+1/m$. For the second parabola $y^2=8(x+2)$, the vertex is at $(-2, 0)$ and $4a=8$, implying $a=2$, so the corresponding tangent equation with slope $m'$ is $y=m'(x+2)+2/m'$.

Since the two tangents are perpendicular, their slopes must satisfy the relation $m' = -1/m$. Substituting this condition into the second tangent equation yields $y = -(1/m)(x+2) - 2m$. To find the intersection point, equate the two expressions for $y$: $mx+m+1/m = -x/m - 2/m - 2m$. Multiplying the entire equation by $m$ eliminates the fractions, resulting in $m^2x + m^2 + 1 = -x - 2 - 2m^2$. Rearranging the terms to group those containing $x$ on one side gives $x(m^2+1) = -3m^2 - 3$. Factoring the right side, we obtain $x(m^2+1) = -3(m^2+1)$. Dividing both sides by $m^2+1$ shows that the intersection point must satisfy the equation $x+3=0$.

For a line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$, the condition $c = a/m$ must hold. Applying this to the parabola $y^2 = 4x$, where $a=1$ and $c=4$, yields $4 = 1/m$, which confirms the slope of the tangent is $m = 1/4$. This establishes the equation of the line as $y = \frac{1}{4}x + 4$.

Substituting this expression for $y$ into the second parabola $x^2 = 2by$ produces the intersection:
$x^2 = 2b\left(\frac{1}{4}x + 4\right)$
This simplifies to the quadratic equation $x^2 - \frac{b}{2}x - 8b = 0$. Because the line is tangent to the parabola, the discriminant of this quadratic equation must be zero. Setting $B^2 - 4AC = 0$ provides:
$\left(-\frac{b}{2}\right)^2 - 4(1)(-8b) = 0$
Expanding this expression gives $\frac{b^2}{4} + 32b = 0$, which is equivalent to $b^2 + 128b = 0$. Factoring the result leads to $b(b + 128) = 0$, and excluding the case where $b=0$, we arrive at $b = -128$.

The latus rectum of the parabola $y^{2}=4x$ is defined by the vertical line $x=1$, which serves as the common chord for both circles. With a total latus rectum length of 4, we can construct a right-angled triangle where the radius $2\sqrt{5}$ is the hypotenuse and half the chord length, 2, is one leg. Applying the Pythagorean theorem to determine the distance from the center of a circle to this common chord, we calculate $\sqrt{(2\sqrt{5})^{2}-2^{2}}$, which simplifies to $\sqrt{20-4}$ and results in 4. Since both circles share this chord and lie on opposite sides, the total distance between their centers is the sum of these two perpendicular distances, which equals 8.

The point where these lines meet the $x$-axis is determined by identifying the specific equation for both the tangent to the curve and the normal to the parabola. The slope of a tangent is found using the derivative of the function at the given coordinate, whereas the slope of a normal is the negative reciprocal of that tangent slope, ensuring the two lines are perpendicular. By deriving these line equations and setting the $y$-value to zero, the intersection point on the $x$-axis can be calculated for both, which allows for the determination of the unknown constant $c$.

For the curve $y = e^{x}$, the derivative is $e^{x}$, establishing the slope at $(c, e^c)$ as $e^c$. The tangent line equation is $y - e^{c} = e^{c}(x - c)$, and setting $y = 0$ results in $-e^{c} = e^{c}(x - c)$, which simplifies to $x = c - 1$. For the parabola $y^{2} = 4x$, implicit differentiation yields $2y \frac{dy}{dx} = 4$, meaning $\frac{dy}{dx} = \frac{2}{y}$. Substituting the point $(1, 2)$ gives a tangent slope of $1$, which means the normal slope is $-1$. The normal line equation is $y - 2 = -1(x - 1)$, or $x + y = 3$, and setting $y = 0$ reveals the $x$-intercept is $3$. Equating the two $x$-intercepts, $c - 1 = 3$ leads to $c = 4$.

A common tangent to two parabolas can be identified by equating their respective tangent conditions, after which we can treat this line as one that must be positioned at a distance $c$ from the origin, where $c$ represents the radius of the circle. A line with slope $m$ acts as a tangent to $y^{2} = 4x$ when it adheres to the form $y = mx + \frac{1}{m}$, whereas it is a tangent to $x^{2} = 4y$ when it follows the form $y = mx - m^{2}$. Because the tangent is shared between the two curves, the $y$-intercepts must be identical, leading to the condition $\frac{1}{m} = -m^{2}$. This simplifies to $m^{3} = -1$, confirming that $m = -1$. Substituting this slope value into the tangent equation provides the common tangent $y = -x - 1$, which is equivalent to $x + y + 1 = 0$. For a circle centered at the origin with radius $c$ to touch this line, the perpendicular distance from the center $(0, 0)$ to the line must equal the radius:
$c = \frac{|0 + 0 + 1|}{\sqrt{1^{2} + 1^{2}}} = \frac{1}{\sqrt{2}}$
Thus, $c$ is equal to $\frac{1}{\sqrt{2}}$.

An equilateral triangle inscribed in a parabola with one vertex at the origin is necessarily symmetric about the parabola's axis, forcing each side emanating from the origin to be inclined at $30^{\circ}$ to that axis. For the parabola $y^2 = 8x$, where the standard form $y^2 = 4ax$ implies $a=2$, points on the curve are represented as $(2t^2, 4t)$. The slope of the line segment from the origin to $(2t^2, 4t)$ must satisfy the condition $\frac{4t}{2t^2} = \tan 30^{\circ}$, which simplifies to $\frac{2}{t} = \frac{1}{\sqrt{3}}$ and leads to $t = 2\sqrt{3}$.

This geometry defines a vertical side length of $8t = 16\sqrt{3}$ between the two points on the parabola. The area is calculated using the formula $\frac{\sqrt{3}}{4} \times (\text{side})^2$, and substituting the side length gives $\frac{\sqrt{3}}{4} \times (16\sqrt{3})^2$, which evaluates to $192\sqrt{3}$ square units.

The slope of a tangent line at any point on a curve is defined by its derivative, which is equivalent to the tangent of the angle $\theta$ that the line makes with the positive x-axis. Differentiating the parabola equation $x^2 = 8y$ with respect to $x$ gives $2x = 8 \frac{dy}{dx}$, which simplifies to $\frac{dy}{dx} = \frac{x}{4}$. By setting this derivative equal to the slope $\tan \theta$, we determine that the x-coordinate of the point of contact is $x = 4 \tan \theta$. Substituting this back into the original equation, we find the corresponding y-coordinate to be $y = \frac{(4 \tan \theta)^2}{8} = 2 \tan^2 \theta$.

Using the point-slope form $y - y_1 = m(x - x_1)$ with the point $(4 \tan \theta, 2 \tan^2 \theta)$ and the slope $m = \tan \theta$, the equation becomes $y - 2 \tan^2 \theta = \tan \theta (x - 4 \tan \theta)$. Expanding the right side yields $y - 2 \tan^2 \theta = x \tan \theta - 4 \tan^2 \theta$. Adding $4 \tan^2 \theta$ to both sides results in $y + 2 \tan^2 \theta = x \tan \theta$. Dividing the entire equation by $\tan \theta$ allows us to isolate $x$, resulting in the final form $x = y \cot \theta + 2 \tan \theta$.

The angle between two intersecting curves is derived from the angle between their tangent lines at that shared point. By setting the expressions $10 - x^2$ and $2 + x^2$ equal to each other, we determine that they intersect where $8 = 2x^2$, which gives $x=2$ (the positive coordinate relevant to the reference point), leading to the intersection point $(2,6)$.

The slopes of the curves at this point are determined by differentiating each function. For $y = 10 - x^2$, the derivative is $\frac{dy}{dx} = -2x$, which equals $-4$ when $x=2$. For $y = 2 + x^2$, the derivative is $\frac{dy}{dx} = 2x$, which equals $4$ at the same coordinate.

The value of $|\tan \theta|$ follows from the geometric formula for the tangent of the angle between two lines with slopes $m_1$ and $m_2$:
$|\tan \theta| = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$
Inserting the slopes of $-4$ and $4$ into this expression yields $\left|\frac{-4 - 4}{1 + (-4)(4)}\right|$, which simplifies to $\left|\frac{-8}{1 - 16}\right|$, resulting in $8/15$.

The center of the circle must lie on the normal line to the parabola at the point $(1, 2)$, ensuring the circle is tangent to the curve at that location. For the parabola $y^{2}=4x$, the derivative at $(1, 2)$ indicates a tangent slope of $1$, implying the normal line has a slope of $-1$ and follows the equation $x+y=3$. Representing the circle with center $(x_1, y_1)$ and radius $r$, the condition that it is tangent to the x-axis means the radius squared is equal to the square of the y-coordinate of the center, or $r^2 = y_1^2$.

Since the circle also passes through $(1, 2)$, substituting these coordinates into the circle equation $(x-x_1)^{2} + (y-y_1)^{2} = y_1^2$ gives $(1-x_1)^{2} + (2-y_1)^{2} = y_1^2$. Expanding and simplifying this relationship leads to $x_1^{2} - 2x_1 - 4y_1 + 5 = 0$. By substituting $y_1 = 3 - x_1$ from the normal equation into this expression, we obtain the quadratic $x_1^{2} + 2x_1 - 7 = 0$, which yields $x_1 = 2\sqrt{2} - 1$ for the smaller circle. Consequently, the radius is $y_1 = 4 - 2\sqrt{2}$, and the area of this circle, defined by $\pi y_1^{2}$, calculates to $8\pi(3-2\sqrt{2})$.

To determine where the tangents intersect, we first identify the points where the line $x-y=3$ meets the curve $y=(x-2)^2-1$. Setting the expressions for $y$ equal gives $x-3=(x-2)^2-1$, which simplifies to $x^2-5x+6=0$, yielding intersection points at $x=2$ and $x=3$, corresponding to coordinates $(2,-1)$ and $(3,0)$ on the curve. Using the derivative $\frac{dy}{dx}=2(x-2)$, the tangent at $x=2$ has a slope of $0$ and an equation of $y=-1$, while the tangent at $x=3$ has a slope of $2$ and an equation of $y=2x-6$. Solving these equations simultaneously, we equate $-1=2x-6$ to find $x=\frac{5}{2}$, resulting in the intersection point $\left(\frac{5}{2}, -1\right)$.

Maximizing the area of an inscribed figure relies on expressing the area as a function of a single variable and finding its critical point using differential calculus. Since the rectangle has its base on the x-axis and two vertices on the parabola $y = 12 - x^2$, we can represent its width as $2x$ and its height as $12 - x^2$, where $x$ is the horizontal distance from the y-axis to the rectangle's vertical side.

The area function for this rectangle is $A = 2x(12 - x^2) = 24x - 2x^3$. To find the maximum area, we determine the derivative $\frac{dA}{dx} = 24 - 6x^2$ and set it equal to zero, which simplifies to $6x^2 = 24$ or $x^2 = 4$. Solving for $x$ gives $x = 2$, and substituting this coordinate back into the area formula results in $2(2)(12 - 4) = 32$ sq. units.

The geometry of a parabola dictates that the triangle formed by two fixed points and a variable point on the intervening arc reaches its maximum area when the tangent line at that variable point is parallel to the fixed chord. For the parabola defined by $y^2 = 4x$, any point on the curve can be parameterized as $(t^2, 2t)$, which maps the given points $A(4, -4)$ and $B(9, 6)$ to $t = -2$ and $t = 3$, respectively.

The area of the triangle formed by these points and a general point $C$ on the arc with parameter $t$ follows the quadratic relationship $5|t^2 - t - 6|$. Maximizing this area is equivalent to finding the vertex of the quadratic function $f(t) = t^2 - t - 6$, which occurs at $t = \frac{1}{2}$. Applying this parameter value to the expression yields $5|\left(\frac{1}{2}\right)^2 - \frac{1}{2} - 6|$, which simplifies to $5|\frac{1}{4} - \frac{2}{4} - \frac{24}{4}|$, ultimately resulting in a maximum area of $31 \frac{1}{4}$ square units.

The geometry of this triangle depends on identifying the vertex and intersection points from the parabolic equation $y^2 = -4(x - a^2)$. The vertex is located at $(a^2, 0)$, and setting $x = 0$ identifies the two points of intersection on the y-axis as $(0, 2a)$ and $(0, -2a)$. These points define a vertical base of length $4a$, while the horizontal distance from the y-axis to the vertex provides a height of $a^2$. Applying the standard formula for the area yields:

$\frac{1}{2} \cdot 4a \cdot a^2 = 250$

This simplifies to $2a^3 = 250$, which leads to $a^3 = 125$ and confirms the value is 5.

The intersection points of a line and a parabola are found by solving their equations simultaneously, which provides the coordinates of the chord's endpoints. Rewriting the line equation $x - \sqrt{2}y + 4\sqrt{2} = 0$ as $y = \frac{x+4\sqrt{2}}{\sqrt{2}}$ and substituting this into $x^2 = 4y$ yields the quadratic equation $x^2 = 2\sqrt{2}x + 16$, which rearranges to $x^2 - 2\sqrt{2}x - 16 = 0$. For this quadratic, the sum of the roots is $x_1 + x_2 = 2\sqrt{2}$ and the product is $x_1x_2 = -16$, allowing the squared difference to be calculated as $(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2 = (2\sqrt{2})^2 - 4(-16) = 8 + 64 = 72$. Since the line has a slope of $m = \frac{1}{\sqrt{2}}$, the difference in vertical coordinates is $y_2 - y_1 = \frac{1}{\sqrt{2}}(x_2 - x_1)$, implying $(y_2 - y_1)^2 = \frac{1}{2}(x_2 - x_1)^2 = \frac{1}{2}(72) = 36$. Combining these values into the distance formula $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ results in $\sqrt{72 + 36}$, which simplifies to $\sqrt{108}$ or $6\sqrt{3}$.

The common tangents satisfy the line equation $y = mx + c$ by simultaneously meeting the tangency conditions for both the parabola $y^2 = 12x$ and the hyperbola $x^2 - y^2/8 = 1$. These conditions, defined by $c = 3/m$ for the parabola and $c^2 = m^2 - 8$ for the hyperbola, combine to form the equation $9/m^2 = m^2 - 8$, which simplifies to the quartic $m^4 - 8m^2 - 9 = 0$. Factoring this expression reveals that $m^2 = 9$, which defines the tangents $y = 3x + 1$ and $y = -3x - 1$, which intersect at the point $P(-1/3, 0)$. Given the hyperbola has foci $S(3, 0)$ and $S'(-3, 0)$, calculating the distances from $P$ results in $SP = 10/3$ and $PS' = 8/3$, identifying the ratio of 5:4.