Parabola – JEE Main MathematicsPractice Questions & PYQs

Points on the parabola $y^2 = 6x$ are defined by parameters $t_1, t_2,$ and $t_3$ with $a = 1.5$, where $A, B,$ and $C$ are represented as $(at_1^2, 2at_1)$, $(at_2^2, 2at_2)$, and $(at_3^2, 2at_3)$ respectively. The line $L$ is $y = 2at_3$, making the perpendicular distances $AM = 2a|t_1 - t_3|$ and $BN = 2a|t_2 - t_3|$. The line $AB$ intersects $L$ at $D$, where the x-coordinate $x_D = a(t_1t_3 + t_2t_3 - t_1t_2)$, resulting in a length $CD = a|(t_1 - t_3)(t_2 - t_3)|$. The ratio simplifies to $\frac{4a^2|t_1 - t_3||t_2 - t_3|}{a|(t_1 - t_3)(t_2 - t_3)|} = 4a$, and squaring this yields $(4a)^2 = 16 \times (1.5)^2 = 36$.

The centroid of triangle $PQR$ is determined by the average of the coordinates of its vertices, where $P(x_1, y_1)$ and $Q(x_2, y_2)$ are intersection points on the parabola $y^2=20x$ and $R$ is the focus $(5, 0)$. Given that the centroid $G$ is at $(10, 10)$, the sum of the y-coordinates must satisfy $(y_1+y_2+0)/3=10$, which implies $y_1+y_2=30$. For the line $y=mx+c$, substituting $x=(y-c)/m$ into the equation of the parabola yields $y^2-(20/m)y+20c/m=0$. Because the sum of the roots $y_1+y_2$ is given by $20/m$, equating this to 30 allows us to determine that $m$ is $2/3$.

Using the condition $c-m=6$, we calculate $c = 2/3 + 6$, which simplifies to $20/3$. Substituting these values into the intersection equation results in $y^2-30y+200=0$, which factors as $(y-10)(y-20)=0$. This provides the y-coordinates of 10 and 20, corresponding to the points $P(5, 10)$ and $Q(20, 20)$ on the parabola. The square of the distance between these points is given by $PQ^2 = (20-5)^2+(20-10)^2$, which simplifies to $15^2+10^2$ and equals 325.

The parabola is defined as the locus of points equidistant from the focus $\left(-\frac{1}{2}, 0\right)$ and the line $y = -\frac{1}{2}$, which results in the equation $\left(x+\frac{1}{2}\right)^2 + y^2 = \left(y+\frac{1}{2}\right)^2$. Expanding both sides and simplifying yields the function $f(x) = x^2+x$. Substituting this into the given expression leads to $\tan^{-1}(\sqrt{x^2+x}) + \sin^{-1}(\sqrt{x^2+x+1}) = \frac{\pi}{2}$. By applying the identity $\tan^{-1}(u) = \cos^{-1}\left(\frac{1}{\sqrt{u^2+1}}\right)$, the expression can be rewritten as $\cos^{-1}\left(\frac{1}{\sqrt{x^2+x+1}}\right) + \sin^{-1}(\sqrt{x^2+x+1}) = \frac{\pi}{2}$. Since $\sin^{-1}(z) + \cos^{-1}(z) = \frac{\pi}{2}$ holds true for any valid input, this equality is satisfied only when the arguments of the inverse functions are identical, implying $\frac{1}{\sqrt{x^2+x+1}} = \sqrt{x^2+x+1}$. Squaring both sides gives $x^2+x+1 = 1$, which simplifies to $x^2+x = 0$. Factoring the quadratic expression $x(x+1) = 0$ reveals exactly two values, $x = 0$ and $x = -1$, which constitute the set $S$.

A line $y = mx + c$ serves as a common tangent to the parabolas $x = 2y^2$ and $x = 1 + y^2$ only under specific geometric conditions. For the first parabola expressed as $y^2 = \frac{1}{2}x$, the standard tangent form is $y = mx + \frac{1}{8m}$ because the coefficient $A$ in $y^2 = 4Ax$ is $\frac{1}{8}$. Substituting this tangent into the second equation $x = 1 + y^2$ leads to a quadratic expression where the discriminant must be zero, which identifies the slope as $m = \frac{1}{2\sqrt{2}}$. This slope yields the tangent equation $x - 2\sqrt{2}y + 1 = 0$. The perpendicular distance from the point $(6, -2\sqrt{2})$ to this line is calculated as $\frac{|6 - 2\sqrt{2}(-2\sqrt{2}) + 1|}{\sqrt{1^2 + (-2\sqrt{2})^2}} = \frac{|6 + 8 + 1|}{\sqrt{9}} = 5$.

The point on the parabola $x=4y^2$ that is closest to a given external point $Q(0,33)$ corresponds to the foot of the normal line passing through that point. By rewriting the equation as $y^2 = \frac{1}{4}x$, we identify the parameter $a = \frac{1}{16}$. Using the normal equation $y = -tx + 2at + at^3$, we substitute the point $(0,33)$ into the equation to get $33 = \frac{t}{8} + \frac{t^3}{16}$, which simplifies to $t^3 + 2t - 528 = 0$. The real root of this equation is $t=8$, which gives the coordinates of $P$ as $(at^2, 2at) = (\frac{64}{16}, \frac{16}{16}) = (4,1)$.

To find the distance from $P$ to the directrix of the second parabola $y^2 = 4(x+y)$, we complete the square to transform the equation into standard form, resulting in $(y-2)^2 = 4(x+1)$. This parabola has a focal length $a=1$ and a vertex at $(-1,2)$, which places the directrix at $x+1 = -a$, or $x = -2$. The perpendicular distance from the point $(4,1)$ to the vertical line $x = -2$ is given by the difference in their x-coordinates, $|4 - (-2)|$, which equals 6.

The condition that the tangent to a curve at a specific point is perpendicular to a given line relies on the product of their slopes being negative one. By differentiating the parabola $y^2 = 12x$, we find the derivative $\frac{dy}{dx} = \frac{6}{y}$, which gives a slope of $\frac{6}{\alpha}$ at the point $(3, \alpha)$. Given that this tangent is perpendicular to $2x + 2y = 3$, which has a slope of $-1$, we set the product of the slopes $\frac{6}{\alpha} \cdot (-1) = -1$, which yields $\alpha = 6$. Substituting this value into the hyperbola equation $\alpha^2 x^2 - 9y^2 = 9\alpha^2$ results in $36x^2 - 9y^2 = 324$, which simplifies to the standard form $\frac{x^2}{9} - \frac{y^2}{36} = 1$.

With the hyperbola defined, the point on the curve is $(5, 8)$. Applying the standard formula for a normal to a hyperbola $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$, we obtain the equation $\frac{9x}{5} + \frac{36y}{8} = 45$, which reduces to $2x + 5y - 50 = 0$. The perpendicular distance from the point $(6, -4)$ to this normal line is calculated using the distance formula $\frac{|2(6) + 5(-4) - 50|}{\sqrt{2^2 + 5^2}} = \frac{58}{\sqrt{29}}$. Squaring this value to find the final result yields 116.

The intersection of the parabolas on the line $y=1$ implies that the quadratic equations $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$ share a common root. Given that $a, b, c$ are in geometric progression, we have $b^2=ac$, so the first equation simplifies to $(\sqrt{a}x+\sqrt{c})^2=0$, which yields the root $x=-\sqrt{\frac{c}{a}}$. Substituting this value into the second equation $dx^2+2ex+f=0$ gives $d\left(\frac{c}{a}\right)-2e\sqrt{\frac{c}{a}}+f=0$. Dividing this result by $c$ leads to $\frac{d}{a}+\frac{f}{c}=2e\frac{\sqrt{c}}{c\sqrt{a}}$, which simplifies to $\frac{d}{a}+\frac{f}{c}=\frac{2e}{\sqrt{ac}}$. Since $b=\sqrt{ac}$, we arrive at $\frac{d}{a}+\frac{f}{c}=2\left(\frac{e}{b}\right)$, proving that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ form an arithmetic progression.

Since the parabola is defined with a focus at $(3,0)$ and a directrix at $x=-3$, it follows the standard form $y^2 = 4ax$ with $a=3$, resulting in the equation $y^2 = 12x$. Any point on this parabola can be expressed parametrically as $(3t^2, 6t)$; if we denote the parameters of points $P$ and $Q$ as $t_1$ and $t_2$ respectively, the condition that their ordinates are in a $3:1$ ratio implies $6t_1 / 6t_2 = 3$, or simply $t_1 = 3t_2$. The point of intersection $R(\alpha, \beta)$ of the tangents at these two parameters is given by the general coordinate formula $(at_1t_2, a(t_1+t_2))$.

Substituting the known values $a=3$ and $t_1=3t_2$ into these formulas gives $\alpha = 3(3t_2 \cdot t_2) = 9t_2^2$ and $\beta = 3(3t_2 + t_2) = 12t_2$. To find the required ratio, we evaluate $\frac{\beta^2}{\alpha}$ as $\frac{(12t_2)^2}{9t_2^2}$, which simplifies to $\frac{144t_2^2}{9t_2^2} = 16$.

The condition for a line to be tangent to both the parabola $y^2=16x$ and the circle $x^2+y^2=8$ requires the general tangent line $y=mx+\frac{4}{m}$ to maintain a perpendicular distance from the origin equal to the circle's radius $2\sqrt{2}$. Setting up this condition as
$\frac{\left|\frac{4}{m}\right|}{\sqrt{m^2+1}} = 2\sqrt{2}$
simplifies to $m^2=1$, and choosing $m=1$ defines the common tangent line $y=x+4$. The intersection point $R$ where this line touches the parabola is found at $(\frac{4}{1^2}, \frac{2(4)}{1}) = (4,8)$, while the corresponding point of contact $Q$ where the line touches the circle is $(-2,2)$. Consequently, the squared distance $(QR)^2$ is calculated as $(4 - (-2))^2 + (8 - 2)^2$, which equals $36 + 36 = 72$.

To identify the properties of the parabola $y^2 = 8x + 4y + 4$, first rearrange the equation into standard form by completing the square for the $y$ terms, resulting in $(y-2)^2 = 8(x+1)$. This reveals a parabola opening to the right with its vertex at $(-1, 2)$ and a parameter $4a = 8$, which gives $a = 2$. Based on the standard configuration $(y-k)^2 = 4a(x-h)$, the focus is located at the point $(h+a, k)$, which corresponds to $(-1+2, 2)$, or $(1, 2)$.

Because the focal chord passes through this focus $(1, 2)$ and is given to have an $x$-intercept at $(3, 0)$, the slope $m$ of the line connecting these two points is calculated as $\frac{2-0}{1-3} = -1$. The length of a focal chord for a parabola with slope $m$ is determined by the formula $4a(1 + \frac{1}{m^2})$. Substituting $a=2$ and $m=-1$ into this expression yields $4(2)(1 + \frac{1}{(-1)^2})$, which simplifies to $8(1+1)$ for a final length of 16.

The points of tangency on a curve are determined by matching the curve's derivative to the specified tangent slope. For the parabola $y^2 = 3x$, we find the derivative $2y(dy/dx) = 3$, which gives a slope of $3/(2y)$. Since the tangent is parallel to the line $x + 2y = 1$, which has a slope of $-\frac{1}{2}$, we equate $\frac{3}{2y} = -\frac{1}{2}$ to find that $y = -3$. Substituting this value back into the parabola equation yields $x = 3$, identifying point $P$ as $(3, -3)$.

For the ellipse $\frac{x^2}{4} + y^2 = 1$, the tangents at points $Q$ and $R$ must be perpendicular to $x - y = 2$. Given that this line has a slope of $1$, the perpendicular tangent must have a slope of $-1$. Differentiating the ellipse equation leads to $\frac{2x}{4} + 2y(dy/dx) = 0$, or $dy/dx = -\frac{x}{4y}$. Setting this derivative equal to $-1$ results in the relationship $x = 4y$. Substituting $x = 4y$ into the ellipse equation produces $4y^2 + y^2 = 1$, which simplifies to $y = \pm \frac{1}{\sqrt{5}}$, and consequently $x = \pm \frac{4}{\sqrt{5}}$. This establishes the coordinates for $Q$ and $R$ as $(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$ and $(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}})$.

The area of triangle $PQR$ is calculated using the vertices $(3, -3)$, $(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$, and $(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}})$. Applying the standard coordinate geometry formula for the area of a triangle, we have:

$\text{Area} = \frac{1}{2} \left| 3 \left( \frac{1}{\sqrt{5}} - \left( -\frac{1}{\sqrt{5}} \right) \right) + \frac{4}{\sqrt{5}} \left( -\frac{1}{\sqrt{5}} - (-3) \right) - \frac{4}{\sqrt{5}} \left( -3 - \frac{1}{\sqrt{5}} \right) \right|$

Evaluating this expression simplifies the area to $3\sqrt{5}$.

The lines $x=0$ and $y=3$ are perpendicular and intersect at the point $(0, 3)$, which means any triangle formed by these two lines and a third variable line will always be a right-angled triangle. For any right-angled triangle, the circumcentre is simply the midpoint of the hypotenuse, which in this case is the segment of the variable line lying between the two axes.

The variable line is a tangent to the parabola $y^2=6x$, having the equation $y=mx+\frac{3}{2m}$ because $a=\frac{3}{2}$ in the general form $y=mx+\frac{a}{m}$. The intersections of this tangent line with $x=0$ and $y=3$ are the points $(0, \frac{3}{2m})$ and $(\frac{6m-3}{2m^2}, 3)$, respectively. Calculating the midpoint $(h, k)$ of these two points yields $h = \frac{6m-3}{4m^2}$ and $k = \frac{6m+3}{4m}$. By expressing $m$ in terms of $k$ from the second equation and substituting it into the first, we eliminate the parameter $m$ to obtain the relationship $3h = 2(-2k^2+9k-9)$, which simplifies to the locus equation $4y^2-18y+3x+18=0$.

Since the point $P(b, c)$ lies on the parabola $y^2 = 2ax$, the condition $c^2 = 2ab$ must hold. The equation for the tangent line at $P(b, c)$ is expressed as $cy = a(x + b)$. To find the vertices of the triangle formed by this tangent, the vertical line $x = b$, and the x-axis ($y = 0$), we determine the intersection points of these lines. Setting $y = 0$ in the tangent equation gives $x = -b$, providing the vertex $(-b, 0)$. Setting $x = b$ in the tangent equation gives $cy = a(2b) = 2ab = c^2$, which simplifies to $y = c$, placing the vertex at $(b, c)$. The intersection of $x = b$ and $y = 0$ provides the final vertex at $(b, 0)$. Consequently, the triangle is a right-angled triangle with a base along the x-axis of length $2b$ and a height of $c$, resulting in an area calculation of $\frac{1}{2} \times 2b \times c = bc$.

Given that this area equals 16 square units, we have the relation $bc = 16$. With $b$ and $c$ restricted to natural numbers, the possible pairs $(b, c)$ are $(1, 16), (2, 8), (4, 4), (8, 2),$ and $(16, 1)$. For each pair, we calculate $a$ using the relation $a = \frac{c^2}{2b}$ to check if it results in a natural number. For $(1, 16)$, $a = \frac{16^2}{2(1)} = 128$. For $(2, 8)$, $a = \frac{8^2}{2(2)} = 16$. For $(4, 4)$, $a = \frac{4^2}{2(4)} = 2$. The remaining pairs $(8, 2)$ and $(16, 1)$ yield $a = \frac{4}{16} = 0.25$ and $a = \frac{1}{32} = 0.03125$, respectively, which are not natural numbers. Summing the valid values of $a$ belonging to $S$ yields $128 + 16 + 2 = 146$.

The geometric relationship between the parabola $y^2 = 24x$ and the hyperbola $xy = 2$ can be analyzed by considering the tangent to the parabola at a point defined by the parameter $t$, which is given by $ty = x + 6t^2$. When this line intersects the hyperbola at points $A$ and $B$, it serves as a chord with a midpoint $(h, k)$. For the hyperbola $xy = 2$, the equation of a chord with a known midpoint $(h, k)$ is given by the relation $T = S_1$, which results in $kx + hy = 2hk$.

By aligning these two linear equations, we can equate the ratios of their coefficients: $\frac{k}{1} = \frac{h}{-t} = \frac{2hk}{-6t^2}$. From the first ratio, we determine that $t = -h/k$. Substituting this into the second equality, $\frac{h}{-t} = \frac{2hk}{-6t^2}$, simplifies to $1 = \frac{k}{3t}$, or $3t = k$. Replacing $t$ with $-h/k$ leads to the equation $3(-h/k) = k$, which simplifies to $k^2 = -3h$. This indicates that the locus of the midpoint $(h, k)$ is the parabola $y^2 = -3x$. For a parabola of the form $y^2 = -4ax$, the directrix is located at $x = a$. Here, $4a = 3$, so $a = 3/4$, which places the directrix at $x = 3/4$, equivalent to $4x = 3$.

The focal chord length of the parabola $y^2=36x$ is determined by the formula $a(t+1/t)^2=100$, which with $a=9$ simplifies to $9(t+1/t)^2=100$, identifying $t=3$ for point $P(81,54)$ and $Q(1,-6)$. Using the section formula for the ratio $3:1$ to find point $M(21,9)$ on segment $PQ$, we calculate the slope of $PQ$ as $3/4$, establishing that any perpendicular line must have a slope of $-4/3$. The equation of this perpendicular line is derived as $y-9 = -4/3(x-21)$, which simplifies to $4x+3y=111$. Testing the provided coordinates in this equation, the point $(-3,43)$ results in $4(-3)+3(43)=117$, confirming it is the only option that does not lie on the line.

The tangent to a curve at a specific point can be determined by applying standard transformation rules to the curve's equation, replacing quadratic terms and linear variables appropriately based on the coordinates of the point $P(1,3)$. For the curve $x^2 + 2x - 4y + 9 = 0$, substituting these values yields the linear equation $1(x) + (x+1) - 2(y+3) + 9 = 0$, which simplifies to $2x - 2y + 4 = 0$, or $x - y + 2 = 0$. To find the coordinates of $A$, we determine where this line intersects the $y$-axis by setting $x = 0$, resulting in $A(0,2)$.

The line passing through $P(1,3)$ that is parallel to $x - 3y = 6$ must possess the same slope, $1/3$, and its equation can be expressed as $y - 3 = \frac{1}{3}(x - 1)$, which simplifies to $x - 3y + 8 = 0$, or $x = 3y - 8$. Substituting this into the parabola $y^2 = 4x$ creates a quadratic equation $y^2 = 4(3y - 8)$, which simplifies to $y^2 - 12y + 32 = 0$. The roots of this quadratic, $(y-4)(y-8) = 0$, indicate that the line intersects the parabola at points with $y$-coordinates $4$ and $8$, corresponding to points $(4,4)$ and $(16,8)$. Since point $B$ must satisfy the condition $2x - 3y = 8$, we test both coordinates and identify that $(16,8)$ satisfies the equation since $2(16) - 3(8) = 32 - 24 = 8$, whereas $(4,4)$ does not.

With points $A(0,2)$ and $B(16,8)$ identified, calculating the square of the distance between them follows from the standard distance formula:
$(AB)^2 = (16 - 0)^2 + (8 - 2)^2 = 16^2 + 6^2 = 256 + 36 = 292$

Vertex $A$ lies on the $y$-axis, which means its $x$-coordinate is $0$. Evaluating the given line equations at $x=0$ gives $y = \frac{4}{\lambda}$ and $y = \frac{\lambda}{\lambda-1}$ for lines $AB$ and $AC$ respectively. Equating these two expressions allows us to solve for $\lambda$, yielding $\lambda = 2$, which identifies $A$ as $(0, 2)$ and defines the side equations as:
$AB: 3x + 2y = 4, \quad AC: 2x - y + 2 = 0$

To find vertex $C$, we represent it as $(\alpha, 2\alpha + 2)$ to satisfy the equation of line $AC$. Because the altitude from $C$ must pass through the orthocentre $(1, 2)$ and remain perpendicular to $AB$, its slope must be the negative reciprocal of the slope of $AB$. Since the slope of $AB$ is $-\frac{3}{2}$, the slope of the altitude must be $\frac{2}{3}$. Setting the slope between $C$ and the orthocentre equal to $\frac{2}{3}$ allows us to solve for $\alpha$, revealing $\alpha = -\frac{1}{2}$, which places $C$ at $(-\frac{1}{2}, 1)$.

Finally, the tangent from $C$ to the parabola $y^2 = 6x$ follows the form $y = mx + \frac{3}{2m}$. Substituting the coordinates of $C$ into this equation results in $m^2 + 2m - 3 = 0$, yielding slopes $m=1$ and $m=-3$. Testing these confirms that $m=1$ produces a tangent point of $(1.5, 3)$ in the first quadrant. Calculating the distance between point $C(-\frac{1}{2}, 1)$ and $(1.5, 3)$ results in $\sqrt{(1.5 - (-0.5))^2 + (3 - 1)^2} = \sqrt{2^2 + 2^2} = 2\sqrt{2}$.

To find the length of the common tangent segment, we first identify the general equation for a line tangent to the parabola $y^2=4x$, which can be expressed as $y=mx+\frac{1}{m}$. For this line to also be a tangent to the circle $(x-4)^2+y^2=16$, the perpendicular distance from the center of the circle, $(4, 0)$, to the line $mx-y+\frac{1}{m}=0$ must equal the radius of the circle, which is $4$. Setting up this distance condition gives the equation $\left| \frac{4m+\frac{1}{m}}{\sqrt{1+m^2}} \right| = 4$.

Squaring both sides of this equation yields $(4m+\frac{1}{m})^2 = 16(1+m^2)$, which simplifies to $16m^2+8+\frac{1}{m^2} = 16+16m^2$. By canceling the $16m^2$ terms, we find that $\frac{1}{m^2} = 8$, or $m^2=\frac{1}{8}$. The point of contact $P$ on the parabola $y^2=4x$ for a given slope $m$ is defined by the coordinates $(\frac{1}{m^2}, \frac{2}{m})$. Substituting $m^2=\frac{1}{8}$, the $x$-coordinate becomes $8$ and the $y$-coordinate becomes $4\sqrt{2}$.

Because $Q$ is the point of contact on the circle, the length of the tangent segment $PQ$ is simply the distance from point $P$ to the circle along the common tangent. The squared length of the tangent segment from an external point to a circle is equal to the power of the point with respect to that circle. Evaluating the expression $(x-4)^2+y^2-16$ at the coordinates $(8, 4\sqrt{2})$ provides the value of $(PQ)^2$, calculated as $(8-4)^2+(4\sqrt{2})^2-16$, which results in $16+32-16$ for a final value of $32$.