Parabola – JEE Main MathematicsPractice Questions & PYQs

For a parabola $y^2 = 4ax$, any chord connecting points $P_1(at_1^2, 2at_1)$ and $P_2(at_2^2, 2at_2)$ subtends a right angle at the vertex if the product of their parameters satisfies $t_1 t_2 = -4$. Applying this to the equation $y^2 = 12x$ where $4a = 12$, we find $a = 3$, meaning the slopes of the segments connecting the vertex to $P_1$ and $P_2$ are $\frac{2}{t_1}$ and $\frac{2}{t_2}$ respectively, which confirm the condition $t_1 t_2 = -4$ since the lines are perpendicular. Substituting the parametric coordinates into the expression $x_1 x_2 - y_1 y_2$ yields $(3t_1^2)(3t_2^2) - (6t_1)(6t_2)$, which simplifies to $9(t_1 t_2)^2 - 36(t_1 t_2)$. Plugging $t_1 t_2 = -4$ into this expression leads to $9(-4)^2 - 36(-4)$, resulting in a final value of $144 + 144 = 288$.

Since the vertex $O$ of the parabola $x^2 = 4y$ is at the origin $(0,0)$, any point $Q$ on this curve can be expressed in terms of a parameter $t$ as $(2t, t^2)$. A point $P(x,y)$ that divides the segment $OQ$ internally in the ratio $2:3$ satisfies the section formula, yielding coordinates $x = \frac{2(2t) + 3(0)}{5} = \frac{4t}{5}$ and $y = \frac{2(t^2) + 3(0)}{5} = \frac{2t^2}{5}$. By solving for $t = \frac{5x}{4}$ and substituting this into the equation for $y$, the locus of $P$, denoted as conic $C$, simplifies to $5x^2 = 8y$.

For any conic represented by $S(x,y) = 5x^2 - 8y = 0$, the equation of a chord bisected at a specific midpoint $(x_1, y_1)$ is found using the property $T = S_1$. Substituting the point $(1,2)$ into the conic equation yields $S_1 = 5(1)^2 - 8(2) = 5 - 16 = -11$. The expression $T$ is obtained by replacing $x^2$ with $x(1)$ and $y$ with $\frac{y+2}{2}$, resulting in $T = 5x - 8\left(\frac{y+2}{2}\right) = 5x - 4y - 8$. Equating $T$ to $S_1$ results in the equation $5x - 4y - 8 = -11$, which rearranges to:
$5x - 4y + 3 = 0$

The geometry of a focal chord on a parabola $y^2 = 4ax$ relies on the property that the parameters $t_1$ and $t_2$ of its endpoints satisfy $t_1 t_2 = -1$. Given the parabola $y^2 = 16x$, we determine $a = 4$. The endpoint $A(16, 16)$ satisfies the parametric equations $x = at^2$ and $y = 2at$, leading to a parameter $t_1 = 2$. Consequently, the other endpoint $B$ possesses a parameter $t_2 = -1/2$, resulting in the coordinates $(1, -4)$.

When a point $P(\alpha, \beta)$ divides the segment $AB$ in a ratio of $5:2$, the section formula provides two possible coordinates depending on the order of the division. Applying the formula $(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n})$, if $P$ divides the chord in a $5:2$ ratio relative to $A$, the coordinates are $(\frac{5(1) + 2(16)}{7}, \frac{5(-4) + 2(16)}{7}) = (\frac{37}{7}, \frac{12}{7})$, yielding $\alpha + \beta = 7$. Alternatively, if $P$ divides the chord in a $2:5$ ratio, the coordinates become $(\frac{2(1) + 5(16)}{7}, \frac{2(-4) + 5(16)}{7}) = (\frac{82}{7}, \frac{72}{7})$, which results in $\alpha + \beta = 22$. Comparing these values, the minimum value is $7$.

The relationship between the roots of a quadratic equation $ax^2 + bx + c = 0$ is governed by Vieta's formulas, which state that the sum of the roots is $-b/a$ and their product is $c/a$. Representing the roots as $\alpha$ and $2\alpha$, their sum is $3\alpha = \frac{-9(k-1)}{k^2 - 15k + 27}$, which simplifies to $\alpha = \frac{-3(k-1)}{k^2 - 15k + 27}$. Simultaneously, the product of the roots is given by $2\alpha^2 = \frac{18}{k^2 - 15k + 27}$, which implies $\alpha^2 = \frac{9}{k^2 - 15k + 27}$. By squaring the expression for $\alpha$ derived from the sum and equating it to the expression for $\alpha^2$ from the product, we obtain $\frac{9(k-1)^2}{(k^2 - 15k + 27)^2} = \frac{9}{k^2 - 15k + 27}$. Canceling the common factors reveals that $(k-1)^2 = k^2 - 15k + 27$, which expands to $k^2 - 2k + 1 = k^2 - 15k + 27$. Solving for $k$ yields $13k = 26$, resulting in $k = 2$. Substituting this value into the given equation of the parabola $y^2 = 6kx$ produces $y^2 = 12x$. Comparing this to the standard form $y^2 = 4ax$, where $4a$ is the length of the latus rectum, we identify that the length is 12.

Connecting the origin $(0,0)$ to points $P(t_1^2, 2t_1)$ and $Q(t_2^2, 2t_2)$ on the parabola $y^2 = 4x$ results in slopes of $\frac{2}{t_1}$ and $\frac{2}{t_2}$ for the chords. Since the chords are perpendicular, the product of their slopes is $-1$, which simplifies to the condition $t_1 t_2 = -4$.

For the midpoint $(x, y)$ of the segment $PQ$, the coordinates are given by $x = \frac{t_1^2 + t_2^2}{2}$ and $y = t_1 + t_2$. By squaring the expression for $y$, we get $y^2 = t_1^2 + t_2^2 + 2t_1 t_2$, which, upon substituting $t_1^2 + t_2^2 = 2x$ and $t_1 t_2 = -4$, transforms into $y^2 = 2x - 8$. This simplifies to $y^2 = 2(x - 4)$, representing a parabola where the length of the latus rectum, defined by the coefficient of the linear term, is $2$.

Reflecting the parabola $x^2=4y$ across the line $x-y-1=0$ involves transforming a general parametric point $(2t, t^2)$ on the original curve to a new coordinate $(x, y)$. Using the reflection formula $\frac{x-h}{A} = \frac{y-k}{B} = -2\frac{Ah+Bk+C}{A^2+B^2}$ with $A=1, B=-1, C=-1, h=2t,$ and $k=t^2$, the relationship between the reflected coordinates and the parameter $t$ becomes $\frac{x-2t}{1} = \frac{y-t^2}{-1} = -(2t-t^2-1)$. Separating these equations reveals $x = t^2+1$ and $y = 2t-1$. Solving the second equation for $t$ gives $t = \frac{y+1}{2}$, and substituting this into the expression for $x$ yields $x = (\frac{y+1}{2})^2 + 1$. Rearranging this equation into the target form results in $(y+1)^2 = 4(x-1)$. By comparing this with the provided expression $(y+a)^2 = b(x-c)$, we identify $a=1, b=4,$ and $c=1$, which produce a sum of $1+4+1 = 6$.

Any point on the parabola $y^2 = 16x$ can be expressed parametrically as $(4t^2, 8t)$. Since vertex $B$ is $(4, 8)$, it corresponds to the parameter $t=1$. Designating the other two vertices $A$ and $C$ with parameters $t_1$ and $t_2$, the condition that the triangle is right-angled at $B$ requires the product of the slopes $m_{AB}$ and $m_{BC}$ to be $-1$. Calculating these slopes as $\frac{8t_1-8}{4t_1^2-4}$ and $\frac{8t_2-8}{4t_2^2-4}$ leads to the requirement $\frac{2}{t_1+1} \cdot \frac{2}{t_2+1} = -1$, which simplifies to the relation $t_1t_2 + t_1 + t_2 = -5$.

The centroid $(h, k)$ of triangle $ABC$ is determined by the coordinates $(\frac{4t_1^2+4t_2^2+4}{3}, \frac{8t_1+8t_2+8}{3})$. From the equation for $k$, we find $t_1+t_2 = \frac{3k}{8} - 1$. Substituting $t_1t_2 = -5 - (t_1+t_2)$ into the expression for $h$ using the identity $t_1^2 + t_2^2 = (t_1+t_2)^2 - 2t_1t_2$ yields the locus of the centroid:
$h = \frac{9}{48}k^2 + \frac{40}{3}$
This locus equation, which corresponds to the form $x = Ay^2 + C$, indicates a latus rectum of length $\frac{1}{A} = \frac{48}{9} = \frac{16}{3}$. Consequently, three times the length of the latus rectum is $3 \times \frac{16}{3}$, which results in $16$.

The locus of the midpoint of any chord of the parabola $y^2 = 4x$ is derived using the standard condition $T = S_1$, where $T$ is the tangent-like equation at the midpoint $(h, k)$ and $S_1$ is the power of the point. For a chord with midpoint $(h, k)$ in the parabola $y^2 = 4x$, the equation is $ky - 2(x + h) = k^2 - 4h$. Since the chord passes through the origin $(0, 0)$, substituting these coordinates results in $-2h = k^2 - 4h$, which simplifies to $k^2 = 2h$. Consequently, the locus $S$ of all such midpoints is defined by the curve $y^2 = 2x$.

Representing a point $P$ on the curve $S$ using parametric coordinates $\left(\frac{t^2}{2}, t\right)$ allows for the application of the section formula to find the locus of the point $Q(x, y)$ that divides the segment $OP$ internally in a $3:1$ ratio. The coordinates of $Q$ are determined by $\left( \frac{3 \cdot \frac{t^2}{2} + 0}{4}, \frac{3 \cdot t + 0}{4} \right)$, yielding $x = \frac{3t^2}{8}$ and $y = \frac{3t}{4}$. Solving for $t$ gives $t = \frac{4y}{3}$, and substituting this value into the expression for $x$ results in $x = \frac{3}{8} \cdot \left( \frac{16y^2}{9} \right)$, which simplifies to $x = \frac{2y^2}{3}$ or $2y^2 = 3x$.

The vertical distance of a parabola's vertex from the x-axis is defined by the absolute value of the y-coordinate of the vertex. For the equation $y = x^{2} + px + q$, the y-coordinate of the vertex is given by $y_{v} = q - \frac{p^{2}}{4}$. Because the parabola passes through the point $(1, -1)$, we must satisfy the condition $-1 = (1)^{2} + p(1) + q$, which simplifies to the relationship $q = -2 - p$.

Substituting this expression for $q$ into the formula for the vertex height, we get $y_{v} = -2 - p - \frac{p^{2}}{4}$. Factoring out the negative sign, this becomes $y_{v} = -\frac{1}{4}(p^{2} + 4p + 8)$, which can be further rewritten by completing the square as $y_{v} = -\frac{1}{4}((p + 2)^{2} + 4)$. The distance from the x-axis is the magnitude of this value, $|y_{v}| = \frac{1}{4}(p + 2)^{2} + 1$.

To minimize this distance, the quadratic term must be as small as possible, which occurs when $(p + 2)^{2} = 0$, or $p = -2$. Using the earlier relation $q = -2 - p$, we find that $q = -2 - (-2) = 0$. Consequently, the value of $p^{2} + q^{2}$ is calculated as $(-2)^{2} + 0^{2}$, which equals 4.

The parabola defined by $y^2=8x$ corresponds to the standard form $y^2=4ax$, identifying the parameter as $a=2$ and the focus as $A(2, 0)$. Any point on this curve is represented in parametric form as $(2t^2, 4t)$, so we denote points $B$ and $C$ using parameters $t_1$ and $t_2$, respectively. Since the centroid of triangle $ABC$ is $\left(\frac{7}{3}, \frac{4}{3}\right)$, the coordinates of this centroid are the arithmetic means of the coordinates of $A$, $B$, and $C$. Equating the y-coordinate gives $\frac{4t_1 + 4t_2 + 0}{3} = \frac{4}{3}$, which simplifies to $t_1 + t_2 = 1$. Equating the x-coordinate provides $\frac{2t_1^2 + 2t_2^2 + 2}{3} = \frac{7}{3}$, leading to $t_1^2 + t_2^2 = \frac{5}{2}$.

Determining the relationship between the parameters involves the identity $t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2$. Substituting the established values, $\frac{5}{2} = 1^2 - 2t_1t_2$, which yields $2t_1t_2 = -\frac{3}{2}$ or $t_1t_2 = -\frac{3}{4}$. The square of the difference between the parameters is $(t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1t_2 = 1 - 4\left(-\frac{3}{4}\right) = 4$. Using the distance formula, the squared distance between points $B$ and $C$ is $(2t_2^2 - 2t_1^2)^2 + (4t_2 - 4t_1)^2$, which simplifies to $4(t_2 - t_1)^2(t_2 + t_1)^2 + 16(t_2 - t_1)^2$. Substituting the values $4$ and $1$ into this expression results in $4(4)(1) + 16(4) = 16 + 64$, confirming that the squared distance is $80$.

Because triangle OAB is equilateral and the parabola $y^2=4x$ is symmetric about the x-axis, the side OA must be inclined at an angle of $30^\circ$ to the positive x-axis. Using the parametric form for a point on the parabola as $(t^2, 2t)$, the slope of the line segment OA is given by $\frac{2t}{t^2}$, which simplifies to $\frac{2}{t}$. Setting this slope equal to $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ allows us to solve for $t$, resulting in $t = 2\sqrt{3}$. Consequently, point A is located at $(12, 4\sqrt{3})$ and point B is at $(12, -4\sqrt{3})$.

The circle having AB as its diameter possesses a center at the midpoint of the segment, which is $(12, 0)$. The length of the diameter AB is the vertical distance between these two points, $4\sqrt{3} - (-4\sqrt{3}) = 8\sqrt{3}$, meaning the radius of the circle is $4\sqrt{3}$. The minimum distance from the origin to any circle is the difference between the distance from the origin to the center and the radius. Since the origin is at distance $12$ from the center $(12, 0)$, the minimum distance is $12 - 4\sqrt{3}$, which factors to $4(3-\sqrt{3})$.

For the parabola $y^2 = 8x$, the focus lies at $(2, 0)$ and the directrix is the vertical line $x = -2$. The point $A$ where the directrix intersects the $x$-axis is therefore $(-2, 0)$. Any point $B$ on this parabola can be represented in parametric form as $(2t^2, 4t)$ for some parameter $t$. The condition $\alpha > 1$ translates to $2t^2 > 1$, implying $|t| > 1/\sqrt{2}$. Given the slope of the line segment $AB$ is $3/5$, we set up the equation $\frac{4t}{2t^2 + 2} = \frac{3}{5}$. Solving $3t^2 - 10t + 3 = 0$ yields $t=3$ or $t=1/3$. Since $t$ must satisfy the condition $\alpha > 1$, we select $t=3$, which places point $B$ at $(18, 12)$.

Because $BC$ is a focal chord passing through the focus $(2, 0)$, we use the property of parametric coordinates for a focal chord, where the product of the parameters of the endpoints is $-1$. Given the parameter for $B$ is $3$, the parameter for $C$ must be $-1/3$. This provides the coordinates of $C$ as $(2/9, -4/3)$. The area of $\triangle ABC$ with vertices $(-2, 0)$, $(18, 12)$, and $(2/9, -4/3)$ is calculated using the determinant formula, which simplifies to $80/3$. Multiplying this area by $6$ results in $160$.

A parabola defined by $y^2 = 12x$ can be represented using parametric coordinates $P(3t^2, 6t)$, where the vertex $O$ is at the origin. With point $A$ located on the $x$-axis at $(x, 0)$, the geometric requirement that $\angle OPA = 90^{\circ}$ implies that the vectors $\overrightarrow{OP} = (3t^2, 6t)$ and $\overrightarrow{PA} = (x - 3t^2, -6t)$ must be perpendicular. Setting their dot product to zero, $(3t^2)(x - 3t^2) + (6t)(-6t) = 0$, allows us to simplify the expression to $3t^2(x - 3t^2) - 36t^2 = 0$, which yields $x - 3t^2 - 12 = 0$ and confirms that $x = 12 + 3t^2$.

The centroid $(h, k)$ of triangle $OPA$ is calculated by averaging the vertices $(0, 0)$, $(3t^2, 6t)$, and $(12 + 3t^2, 0)$, resulting in $h = \frac{3t^2 + 12 + 3t^2}{3} = 2t^2 + 4$ and $k = \frac{6t}{3} = 2t$. By substituting $t = \frac{k}{2}$ into the expression for $h$, we find $h = 2\left(\frac{k}{2}\right)^2 + 4$, which simplifies to $h = \frac{k^2}{2} + 4$. Multiplying by two and rearranging the terms leads to $k^2 - 2h + 8 = 0$, defining the locus as $y^2 - 2x + 8 = 0$ when expressed in terms of the general coordinates $x$ and $y$.

The angle subtended by a line segment $AB$ at the vertex of a parabola $(0,0)$ is determined by the slopes of the two lines connecting the vertex to these points. If the slopes of these lines are $m_1$ and $m_2$, the tangent of the angle $\theta$ is given by the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.

To identify the coordinates of $A$ and $B$, substitute $y = \frac{3x^2}{4}$ into the line equation $3x - 2y + 12 = 0$. This substitution leads to $3x - 2\left(\frac{3x^2}{4}\right) + 12 = 0$, which simplifies to $3x - \frac{3x^2}{2} + 12 = 0$. Multiplying this equation by $-2$ and dividing by $3$ results in the quadratic equation $x^2 - 2x - 8 = 0$. Factoring this expression gives $(x-4)(x+2) = 0$, identifying the $x$-coordinates as $4$ and $-2$. Calculating the corresponding $y$-values yields the points $A(4, 12)$ and $B(-2, 3)$.

With the vertex at $(0,0)$, the slopes of the lines $OA$ and $OB$ are $m_1 = \frac{12-0}{4-0} = 3$ and $m_2 = \frac{3-0}{-2-0} = -\frac{3}{2}$. Substituting these values into the tangent formula yields $\tan \theta = \left| \frac{3 - (-3/2)}{1 + 3(-3/2)} \right|$. This simplifies to $\left| \frac{4.5}{-3.5} \right|$, which equals $\frac{9}{7}$, making the angle $\tan^{-1}\left(\frac{9}{7}\right)$.

Reflecting an arbitrary point $(t^2, 2t)$ from the parabola $y^2=4x$ across the line $x+y+4=0$ yields coordinates $(x,y)$ related by $x = -2t-4$ and $y = -t^2-4$. Eliminating the parameter $t$ by substituting $t = \frac{-(x+4)}{2}$ into the expression for $y$ produces the equation of the reflected parabola:
$(x+4)^2 = -4(y+4)$
This parabola opens downwards with its vertex located at $(-4, -4)$.

Points $A$ and $B$ are determined by the intersection of this reflected parabola with the line $y = -5$. Substituting $y = -5$ into the equation gives $(x+4)^2 = -4(-5+4)$, which simplifies to $(x+4)^2 = 4$. Solving for $x$ provides $x+4 = 2$ and $x+4 = -2$, resulting in $x=-2$ and $x=-6$. The distance $d$ between $A(-2, -5)$ and $B(-6, -5)$ is the horizontal distance between these two $x$-coordinates, which is $d = 4$.

The area $a$ of $\triangle SAB$ is calculated using the focus $S$ of the original parabola $y^2=4x$, which is $(1, 0)$. Since the segment $AB$ lies on the horizontal line $y = -5$, the base of the triangle is the length of $AB$, which is $4$. The height of the triangle is the vertical distance from $S(1, 0)$ to the line $y = -5$, which is $5$. The area is computed as:
$a = \frac{1}{2} \times 4 \times 5 = 10$
The final sum is given by $a+d = 10 + 4 = 14$.

The equation of the tangent to the parabola $y^2 = x-2$ at a point $B(x_1, y_1)$ is given by $yy_1 = \frac{1}{2}(x+x_1) - 2$. Since this line passes through $A(-2, 0)$, substituting the coordinates gives $0 = \frac{1}{2}(-2+x_1) - 2$, which simplifies to $x_1 = 6$. Substituting $x_1 = 6$ into the parabola equation yields $y_1^2 = 4$, so $y_1 = 2$, identifying the point of tangency as $B(6, 2)$.

The line $AB$ connects $A(-2, 0)$ and $B(6, 2)$, so its slope is $\frac{2-0}{6-(-2)} = \frac{1}{4}$, resulting in the equation $y = \frac{1}{4}(x+2)$, which can be expressed as $x = 4y - 2$. The parabola equation can be expressed as $x = y^2 + 2$. The area of the region bounded by the line, the parabola, and the $x$-axis from $y=0$ to $y=2$ is the integral of the difference between the parabola's $x$-coordinate and the line's $x$-coordinate:
$\int_{0}^{2} ((y^2 + 2) - (4y - 2)) \, dy$
Evaluating this integral gives:
$\int_{0}^{2} (y^2 - 4y + 4) \, dy = \int_{0}^{2} (y-2)^2 \, dy = \left[ \frac{(y-2)^3}{3} \right]_{0}^{2} = 0 - \left( -\frac{8}{3} \right) = \frac{8}{3}$

The harmonic property of parabolic chords passing through the focus states that the sum of the reciprocals of the two segments created by the focus equals the reciprocal of the semi-latus rectum. Given the parabola $y^2 = 4x$, the focus is located at $(1, 0)$ and the value of the semi-latus rectum parameter is $a=1$. For the diagonal $AC$ of length $\frac{25}{4}$ divided into segments $AS=c$ and $CS=\frac{25}{4}-c$, the condition is expressed by the equation:
$\frac{1}{c} + \frac{1}{\frac{25}{4} - c} = 1$
Solving this quadratic relationship, $4c^2 - 25c + 25 = 0$, yields two possible segment lengths: $c = \frac{5}{4}$ and $c = 5$.

The distance from any point $(t^2, 2t)$ on this parabola to the focus is $t^2 + 1$. Equating $t^2 + 1$ to the calculated segment lengths determines the coordinates of the vertices. For $c = \frac{5}{4}$, $t^2 = \frac{1}{4}$ provides the points $(\frac{1}{4}, 1)$ and $(\frac{1}{4}, -1)$, defining a vertical side $AD$ with length 2. For $c = 5$, $t^2 = 4$ leads to the points $(4, 4)$ and $(4, -4)$, which form the parallel vertical side $BC$ with length 8. Since these sides are parallel to the $y$-axis, the height of the trapezium is the distance between the $x$-coordinates $x = \frac{1}{4}$ and $x = 4$, which is $\frac{15}{4}$. The area of the trapezium is calculated as $\frac{1}{2} (2 + 8) \times \frac{15}{4} = \frac{75}{4}$.