Parabola – JEE Main MathematicsPractice Questions & PYQs

The two given parabolas, $y=x^2+2$ and $x=y^2+2$, are perfectly symmetric about the line $y=x$, which implies that the center of the smallest circle touching both must lie on this same line. The circle's diameter is defined by the distance between the two points on the parabolas that are closest to each other, where the tangent slope is equal to $1$ because the normal must be parallel to the line of symmetry $y=x$.

For the curve $x=y^2+2$, taking the derivative with respect to $x$ gives $1 = 2y \frac{dy}{dx}$, so the slope is $\frac{dy}{dx} = \frac{1}{2y}$. Setting this to $1$ finds the tangency at $y = \frac{1}{2}$, yielding an $x$-coordinate of $\frac{9}{4}$, so the contact point is $(\frac{9}{4}, \frac{1}{2})$. Due to symmetry, the contact point on $y=x^2+2$ is the reflection, $(\frac{1}{2}, \frac{9}{4})$. The distance between these two points provides the diameter: $\sqrt{(\frac{9}{4} - \frac{1}{2})^2 + (\frac{1}{2} - \frac{9}{4})^2} = \sqrt{2 \cdot (\frac{7}{4})^2} = \frac{7\sqrt{2}}{4}$. Dividing by $2$, the radius of the circle is $\frac{7\sqrt{2}}{8}$.

To identify the circle's center, rewrite the parabola $y^2 = 4x + 16$ as $y^2 = 4(x + 4)$. This parabola has a vertex at $(-4, 0)$ and a focal length $a=1$, placing its focus at $(-3, 0)$. Because the focus serves as the center of the circle with radius 5, the circle's equation is $(x+3)^2 + y^2 = 25$.

The intersection point of the lines $3x - y = 0$ and $x + \lambda y = 4$ is found by substituting $y = 3x$ into the second equation, yielding $x = \frac{4}{1 + 3\lambda}$ and $y = \frac{12}{1 + 3\lambda}$. Since this point lies on the circle, substituting these coordinates gives $(\frac{4}{1 + 3\lambda} + 3)^2 + (\frac{12}{1 + 3\lambda})^2 = 25$. Simplifying this yields $(\frac{7 + 9\lambda}{1 + 3\lambda})^2 + (\frac{12}{1 + 3\lambda})^2 = 25$, which implies $(7 + 9\lambda)^2 + 144 = 25(1 + 3\lambda)^2$.

Expanding this expression results in the quadratic equation
$6\lambda^2 + \lambda - 7 = 0$
Factoring this gives $(\lambda - 1)(6\lambda + 7) = 0$, resulting in roots $\lambda_1 = -\frac{7}{6}$ and $\lambda_2 = 1$. Consequently, the requested value $12\lambda_1 + 29\lambda_2$ becomes $12(-\frac{7}{6}) + 29(1) = -14 + 29$, which equals 15.

The geometric definition of a parabola states that any point $(x, y)$ on the curve is equidistant from a fixed focus $F$ and a directrix line $L$. Given the axis of the parabola is $y=x$ and the vertex $V$ and focus $F$ are at distances $\sqrt{2}$ and $2\sqrt{2}$ from the origin, their coordinates are $(1, 1)$ and $(2, 2)$ respectively, as they lie on the line $y=x$ in the first quadrant. The vertex is the midpoint between the focus and the intersection point $Z$ of the directrix and the axis; setting the midpoint of $F$ and $Z$ equal to $V$ yields $Z=(0, 0)$. Since the directrix is perpendicular to the axis $y=x$ and passes through the origin, its equation is $x+y=0$.

Applying the condition that the distance from $(x, y)$ to the focus equals the perpendicular distance to the directrix, we write $\sqrt{(x-2)^2 + (y-2)^2} = \frac{|x+y|}{\sqrt{1^2+1^2}}$. Squaring both sides results in $(x-2)^2 + (y-2)^2 = \frac{(x+y)^2}{2}$, which simplifies to $2(x^2 - 4x + 4 + y^2 - 4y + 4) = x^2 + 2xy + y^2$. Expanding and rearranging the terms leads to the equation $(x-y)^2 - 8(x+y) + 16 = 0$. Substituting the point $(1, k)$ into this equation gives $(1-k)^2 - 8(1+k) + 16 = 0$, which simplifies to $k^2 - 2k + 1 - 8 - 8k + 16 = k^2 - 10k + 9 = 0$. Factoring the quadratic yields $(k-9)(k-1) = 0$, resulting in possible values of $k=1$ or $k=9$.

The geometric definition of a parabola states that any point on the curve is equidistant from a fixed focus and a fixed line, the directrix. For a focus at $(4,3)$, the square of the distance to the directrix $x=0$ is $x^2$, while the square of the distance to the directrix $y=0$ is $y^2$. Equating the squared distance from the focus $(4,3)$ to the squared distances from each directrix gives the two equations $(x-4)^2 + (y-3)^2 = x^2$ and $(x-4)^2 + (y-3)^2 = y^2$. These equations imply that at the intersection points, $x^2 = y^2$, which means the intersection must occur along the lines $y=x$ or $y=-x$.

Substituting $y=x$ into the first parabola equation leads to $(x-4)^2 + (x-3)^2 = x^2$. Expanding this expression results in $x^2 - 8x + 16 + x^2 - 6x + 9 = x^2$, which simplifies to the quadratic $x^2 - 14x + 25 = 0$. Because the case $y=-x$ produces a negative discriminant, all real intersection points $A$ and $B$ lie on the line $y=x$. If $x_1$ and $x_2$ are the roots of this quadratic, then $x_1 + x_2 = 14$ and $x_1 x_2 = 25$. The intersection points are $(x_1, x_1)$ and $(x_2, x_2)$, and the squared distance between them is $(AB)^2 = (x_2 - x_1)^2 + (x_2 - x_1)^2 = 2(x_2 - x_1)^2$. Using the identity $(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2$, we calculate $14^2 - 4(25) = 196 - 100 = 96$. Multiplying this by 2 yields a final value of $192$.

A parabola consists of all points equidistant from a focus and a fixed directrix line. Given the vertex $V\left(\frac{3}{2}, 3\right)$ and the directrix $x+2y=0$, we identify the focus as $S(3, 6)$. Establishing the locus equality $PS^2=PM^2$ allows us to define the parabola by equating the squared distance from an arbitrary point $(x, y)$ to the focus and the squared perpendicular distance to the directrix line, expressed as $(x-3)^2 + (y-6)^2 = \left(\frac{x+2y}{\sqrt{5}}\right)^2$. Multiplying both sides by 5 yields $5(x^2 - 6x + 9 + y^2 - 12y + 36) = (x+2y)^2$. Expanding the right side results in $5x^2 - 30x + 45 + 5y^2 - 60y + 180 = x^2 + 4xy + 4y^2$. Rearranging these terms leads to the equation $4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0$. Identifying the coefficients $\alpha = 4$, $\beta = 1$, and $\gamma = 4$ from this form, the sum $\alpha + \beta + \gamma$ results in 9.

The focus of a parabola $y^2 = 4ax$ divides any focal chord $PQ$ in the ratio of the focal distances $PS:SQ$. For a point $P$ defined by the parameter $t$, where $P(at^2, 2at)$, this ratio simplifies to $t^2:1$.

For the parabola $y^2 = 16x$, the parameter $a$ equals $4$. Utilizing the coordinate $2at = -4$ for the given point $P(1, -4)$, we find $8t = -4$, which identifies the parameter as $t = -1/2$. Consequently, the ratio $m:n$ equals $t^2:1$, yielding $(-1/2)^2:1$, or $1:4$. Since $\gcd(1, 4) = 1$, we assign $m=1$ and $n=4$. Calculating the final value $m^2+n^2$ produces $1^2 + 4^2 = 17$.

The family of circles tangent to the parabola $y^2=9x$ at the point $(4, 6)$ can be expressed as $(x-4)^2 + (y-6)^2 + \lambda(3x-4y+12) = 0$, where $3x-4y+12=0$ is the tangent line to the parabola at the contact point. Expanding this expression yields $x^2+y^2+(3\lambda-8)x-(12+4\lambda)y+(52+12\lambda)=0$. For this circle to touch the x-axis, the square of the x-coordinate of the center, $g^2$, must equal the constant term $c$. Using the general circle form coefficients, where $g = \frac{3\lambda-8}{2}$ and $c = 52+12\lambda$, this condition results in the equation $(\frac{3\lambda-8}{2})^2 = 52+12\lambda$.

Simplifying this expression leads to the quadratic equation $9\lambda^2-96\lambda-144=0$, which yields two possible values for $\lambda$, specifically $\lambda=12$ and $\lambda=-\frac{2}{3}$. Because the circle touches the x-axis at $a < 0$, the x-coordinate of the center must be negative, requiring $\lambda > \frac{8}{3}$. This condition uniquely selects $\lambda=12$. Substituting $\lambda=12$ into the expression for the y-coordinate of the center, $f = -(6+2\lambda)$, results in $f = -30$. Since the radius of a circle tangent to the x-axis is given by the absolute value of the center's y-coordinate, $|f|$, the radius $r$ is 30.

The shortest distance from a point on the axis of a parabola to the curve itself is defined along the normal line. Representing any point on the parabola $y^2 = 4x$ as $(t^2, 2t)$, the square of the distance to the point $(a, 0)$ is given by $f(t) = (t^2 - a)^2 + 4t^2$. Minimizing this expression, we set the derivative $f'(t) = 4t^3 - 4t(a - 2) = 0$, which yields the condition $t^2 = a - 2$. Substituting this back into the distance expression, we have $4a - 4 = 16$, which confirms the value of the parameter $a$ is 5.

The circle passes through $(5, 0)$ and the focus of the parabola, which is located at $(1, 0)$. Since the center lies on the axis of the parabola, these two points serve as the endpoints of the diameter of the circle. Applying the diameter form of the circle equation, $(x - 1)(x - 5) + y^2 = 0$, and expanding the terms, we arrive at the equation $x^2 + y^2 - 6x + 5 = 0$.

For the parabola $y^2=12x$, the standard form is $y^2=4ax$, which gives $a=3$ and a focus at $S(3, 0)$. Any point $P$ on this parabola can be expressed in terms of a parameter $t$ as $(3t^2, 6t)$. Because $PQ$ is a focal chord, the coordinates of $Q$ must correspond to the parameter $-1/t$, resulting in $Q\left(\frac{3}{t^2}, -\frac{6}{t}\right)$. The distance from the focus to any point $P(3t^2, 6t)$ is given by $x+a$, yielding $SP = 3t^2+3$ and $SQ = \frac{3}{t^2}+3$. The product of these focal segments is $9(1+t^2)\left(1+\frac{1}{t^2}\right) = 9\frac{(1+t^2)^2}{t^2}$.

Setting this equal to the provided value of $\frac{147}{4}$ gives $\frac{(1+t^2)^2}{t^2} = \frac{49}{12}$. Expanding this equation leads to $12(1+2t^2+t^4) = 49t^2$, which simplifies to the quadratic $12t^4 - 25t^2 + 12 = 0$. Factoring the quadratic results in $(3t^2 - 4)(4t^2 - 3) = 0$, giving $t^2 = \frac{4}{3}$ or $t^2 = \frac{3}{4}$. Choosing $t^2 = \frac{3}{4}$ with $t = -\frac{\sqrt{3}}{2}$ allows us to identify the coordinates of the endpoints as $P\left(\frac{9}{4}, -3\sqrt{3}\right)$ and $Q(4, 4\sqrt{3})$.

The circle $C$ described with $PQ$ as a diameter follows the equation $(x - \frac{9}{4})(x - 4) + (y + 3\sqrt{3})(y - 4\sqrt{3}) = 0$. Expanding the terms produces $x^2 - \frac{25}{4}x + 9 + y^2 - \sqrt{3}y - 36 = 0$, which simplifies to $x^2 + y^2 - \frac{25}{4}x - \sqrt{3}y - 27 = 0$. Multiplying the entire equation by 64 yields $64x^2 + 64y^2 - 400x - 64\sqrt{3}y - 1728 = 0$. Rearranging to the form $64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$ identifies $\alpha = 400$ and $\beta = 1728$. Consequently, $\beta - \alpha = 1728 - 400 = 1328$.

Substituting $P(4, 4\sqrt{3})$ into the parabola equation $y^2=4ax$ yields $(4\sqrt{3})^2=4a(4)$, which simplifies to $48=16a$, establishing that $a=3$ and the parabola is $y^2=12x$ with focus $S(3,0)$ and directrix $x=-3$. The point $M$ on the directrix corresponding to $P$ is $(-3, 4\sqrt{3})$, creating a horizontal segment $PM$ of length $7$. Since $PQ$ is a focal chord passing through $S(3,0)$ and $P(4, 4\sqrt{3})$, its equation is $y=4\sqrt{3}(x-3)$, and solving this simultaneously with $y^2=12x$ identifies $Q$ at $(\frac{9}{4}, -3\sqrt{3})$, which places the foot of the perpendicular $N$ on the directrix at $(-3, -3\sqrt{3})$. The quadrilateral $PQMN$ is a trapezoid with parallel sides $PM=7$ and $QN=\frac{21}{4}$, and a height defined by the vertical distance between $P$ and $Q$ of $7\sqrt{3}$, resulting in an area of

$\frac{1}{2} \times \left(7 + \frac{21}{4}\right) \times 7\sqrt{3} = \frac{343\sqrt{3}}{8}$

The length of a focal chord in a parabola $y^2 = 4ax$ is determined by its inclination angle $\theta$ relative to the axis of the parabola according to the expression $4a \csc^2 \theta$. For the parabola $y^2 = 12x$, identifying the parameter $4a = 12$ places the focus at $(3, 0)$. Setting the length equal to 15 leads to the equation $12 \csc^2 \theta = 15$, which simplifies to $\sin^2 \theta = \frac{4}{5}$ and consequently determines the slope of the chord, $\tan \theta = 2$.

With the slope and the focus $(3, 0)$ known, the equation of the focal chord becomes $y = 2(x-3)$, which rearranges to $2x - y - 6 = 0$. The perpendicular distance $p$ from the origin $(0, 0)$ to this line is found using the standard distance formula, resulting in $p = \frac{|2(0) - 0 - 6|}{\sqrt{2^2 + (-1)^2}} = \frac{6}{\sqrt{5}}$. Squaring this value yields $p^2 = \frac{36}{5}$, making the final calculation for $10 p^2$ equivalent to $10 \cdot \frac{36}{5}$, which simplifies to 72.

Due to the symmetry of the parabola $y=6-x^2$ and the lines $y=\pm\sqrt{3}x$ about the $y$-axis, the center of the circle must lie on the $y$-axis at a coordinate $(0, y_c)$. Since the circle touches the vertex of the parabola at $(0, 6)$, the relationship between its radius $r$ and the center's vertical position is defined by $y_c = 6-r$.

The condition for tangency with the line $\sqrt{3}x - y = 0$ requires that the perpendicular distance from the center $(0, 6-r)$ to the line equals the radius $r$:
$\frac{|-(6-r)|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = r$
Simplifying this expression yields $|r-6| = 2r$, which results in a valid radius of $r=2$. This places the center at $(0, 4)$ and defines the circle by the equation $x^2 + (y-4)^2 = 4$. Testing the point $(2, 4)$ in this equation gives $2^2 + (4-4)^2 = 4$, confirming that it lies on the circle.

The shortest distance from an external point to a parabola is measured along the normal line passing through that point, which allows us to find the distance by identifying the intersection of the normal from the circle's center to the parabola. We first determine the center of the circle $x^2+y^2-4x-16y+64=0$ by completing the squares to get $(x-2)^2 + (y-8)^2 = 4$, placing the center at $(2, 8)$.

For the parabola $y^2=4x$, we have $a=1$, and the equation of a normal with slope $m$ is $y = mx - 2am - am^3$, which simplifies to $y = mx - 2m - m^3$. Since the normal must pass through $(2, 8)$, we substitute these coordinates into the equation to get $8 = 2m - 2m - m^3$, resulting in $m^3 = -8$ and therefore $m = -2$. The point $P$ on the parabola where this normal meets is given by the parametric coordinates $(am^2, -2am)$, which evaluates to $(1(-2)^2, -2(1)(-2)) = (4, 4)$. Finally, the distance $d$ between $(2, 8)$ and $(4, 4)$ is found via the distance formula $d = \sqrt{(4-2)^2 + (4-8)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{20}$, leading to $d^2 = 20$.

The geometry of this conic is determined by the differential relationship between the tangent at any point $P(x, y)$ and the secant line passing through $P$ and the fixed point $(3, -5)$. By setting the slope of the tangent $\frac{dy}{dx}$ equal to half the slope of the line connecting $P$ to $(3, -5)$, we establish the separable differential equation $\frac{dy}{dx} = \frac{1}{2} \left( \frac{y+5}{x-3} \right)$, which rearranges to $\frac{2}{y+5} dy = \frac{1}{x-3} dx$. Integrating both sides results in $2 \ln(y+5) = \ln(x-3) + C$. Substituting the point $(4, -2)$ allows us to calculate $C = 2 \ln(3)$, leading to the parabola equation $(y+5)^2 = 9(x-3)$.

In this parabolic form, the vertex is at $(3, -5)$ and the parameter $4a = 9$, which gives $a = 2.25$. For a parabola of the form $(y-k)^2 = 4a(x-h)$, the focal distance of any point $(x, y)$ is simply $x - h + a$. Evaluating this for the point $(7, 1)$ where $x=7$, $h=3$, and $a=2.25$, we find $d = 7 - 3 + 2.25 = 6.25$, or $d = \frac{25}{4}$. Therefore, the final value of $12 d$ is $12 \times \frac{25}{4} = 75$.

The equation of a chord of a parabola with a given midpoint $(x_1, y_1)$ is determined by the relation $T = S_1$. For the parabola $x^2 = 8y$ and the midpoint $\left(1, \frac{5}{4}\right)$, this yields the equation $x(1) - 4(y + \frac{5}{4}) = 1^2 - 8(\frac{5}{4})$. This simplifies to $x - 4y - 5 = 1 - 10$, which results in the linear relationship $x - 4y + 4 = 0$.

Substituting the coordinates of point $P(\alpha, \beta)$ into this chord equation gives $\alpha - 4\beta + 4 = 0$, implying $\alpha = 4\beta - 4$. Since $P$ also lies on the parabola $y^2 = 4x$, replacing $x$ with $4\beta - 4$ produces the quadratic equation $\beta^2 = 4(4\beta - 4)$, which simplifies to $\beta^2 - 16\beta + 16 = 0$.

Calculating the value of $(\alpha - 28)(\beta - 8)$ involves substituting $\alpha = 4\beta - 4$ into the expression, transforming it into $(4\beta - 4 - 28)(\beta - 8)$. This simplifies to $(4\beta - 32)(\beta - 8)$, or $4(\beta - 8)^2$. Expanding $(\beta - 8)^2$ results in $\beta^2 - 16\beta + 64$. From the quadratic $\beta^2 - 16\beta + 16 = 0$, we know that $\beta^2 - 16\beta = -16$. Substituting this into the expression leads to $4 \times (-16 + 64)$, which equals $4 \times 48$, resulting in 192.

The focal chord of the parabola $y^2=12x$ has a length $l = 4a \operatorname{cosec}^2 \theta$, where $a=3$ and $\theta$ is the angle the chord makes with the axis. A chord with slope $m = \tan \theta$ passing through the focus $(3,0)$ follows the equation $mx - y - 3m = 0$, making the perpendicular distance $d$ from the origin to this line equal to $\frac{3|m|}{\sqrt{m^2+1}}$. By using the trigonometric identity $\sin \theta = \frac{|m|}{\sqrt{1+m^2}}$, the distance simplifies to $d = 3 \sin \theta$. Multiplying the length $l = \frac{12}{\sin^2 \theta}$ by the square of this distance $d^2 = 9 \sin^2 \theta$ cancels the trigonometric terms, resulting in the constant value $l d^2 = 108$.

The equation of a chord of a parabola with a given midpoint is determined by the formula $T = S_1$, where $T$ represents the expression $yy_1 - 2a(x+x_1)$ and $S_1$ is the value $y_1^2 - 4ax_1$. For the parabola $y^2 = 12x$, we have $4a = 12$, which implies $2a = 6$; substituting the midpoint $(4, 1)$ into this relation yields $y(1) - 6(x + 4) = 1^2 - 12(4)$. Simplifying the left side results in $y - 6x - 24$ and the right side evaluates to $-47$, leading to the linear equation $y - 6x = -23$, or equivalently, $6x - y = 23$. Testing the coordinates from the options against this equation reveals that only the point $(\frac{1}{2}, -20)$ satisfies the condition, as $6(\frac{1}{2}) - (-20) = 3 + 20 = 23$.

The slope of the line $2x-y=10$ is $2$, so any perpendicular tangent line must have a slope of $m = -1/2$. For the parabola $y^2 = 4(x-9)$, which takes the form $y^2 = 4a(x-h)$ with $a=1$ and $h=9$, the point of contact $P$ for a given slope $m$ is found at $(h+a/m^2, 2a/m)$. Substituting the known values leads to the coordinates $P(13, -4)$.

Completing the square for the circle equation $x^2+y^2-14x-8y+56=0$ reveals it can be written as $(x-7)^2 + (y-4)^2 = 9$, which identifies the center of the circle as $(7, 4)$. The distance $d$ between this center and the point $P(13, -4)$ is calculated as follows:
$d = \sqrt{(13-7)^2 + (-4-4)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36+64} = 10$

The parabola equation $9 x^2+12 x+18 y-14=0$ can be rearranged into the standard form $(3 x+2)^2=-18(y-1)$. For lines $y=m x+1$ passing through $(0,1)$ to be tangent, the intersection equation $9 x^2+(12+18 m) x+4=0$ must satisfy the condition for a zero discriminant. Setting $(12+18 m)^2-144=0$ yields $12+18 m=\pm 12$, which produces the two tangent slopes $m=0$ and $m=-4/3$.

In the isosceles triangle $\triangle PQR$ with base $QR$, the altitude from $P$ coincides with the angle bisector of $\angle QPR$, which implies the base $QR$ is perpendicular to this bisector. Denoting the angle of the tangents as $\theta$ where $\tan \theta=-4/3$, the slope of $QR$ is $m_{QR} = -\cot (\theta/2)$. Utilizing the tangent double-angle formula $\frac{2 \tan (\theta/2)}{1-\tan ^2 (\theta/2)}=-4/3$, we solve the quadratic equation $2 \tan ^2 (\theta/2)-3 \tan (\theta/2)-2=0$ to find the values $\tan (\theta/2)=2$ and $\tan (\theta/2)=-1/2$. Consequently, the possible slopes for $QR$ are $m_1=-1/2$ and $m_2=2$, and the final calculation of $16(m_1^2+m_2^2)$ becomes $16(1/4+4)$, which equals $68$.