Parabola – JEE Main MathematicsPractice Questions & PYQs

The orthocentre of a triangle is the point where its altitudes intersect, but for a right-angled triangle, this point simplifies to the vertex containing the right angle. We begin by identifying the chord of contact for the parabola $y^2 = 2x - 3$ from the point $R(0, 1)$, which is given by the equation $x - y = 3$. Solving this equation simultaneously with the parabola yields the coordinates of points $P$ and $Q$. Substituting $x = y + 3$ into the parabola's equation results in $y^2 = 2(y + 3) - 3$, or $y^2 - 2y - 3 = 0$, which factors to $(y - 3)(y + 1) = 0$, giving us $y = 3$ and $y = -1$. This corresponds to the points $P(6, 3)$ and $Q(2, -1)$.

Calculating the slopes of segments $PQ$ and $RQ$ reveals that the slope of $PQ$ is $1$ and the slope of $RQ$ is $-1$. Since their product is $-1$, the lines are perpendicular, confirming that $\angle PQR = 90^{\circ}$. Because the right angle is at $Q$, the orthocentre of triangle $PQR$ is simply the coordinates of $Q$, which are $(2, -1)$.

For the parabola $y^2 = 4x$, the value of the parameter $a$ is 1. Points on this parabola are represented by the coordinates $(t^2, 2t)$, and for any focal chord $PQ$ with endpoints $P(t_1^2, 2t_1)$ and $Q(t_2^2, 2t_2)$, the relationship $t_1 t_2 = -1$ must hold. The length of such a focal chord is given by the formula $a(t_1 - t_2)^2$, which simplifies to $(t_1 - t_2)^2$ since $a$ is 1.

Given that the length of the chord is 6.25, the equation $(t_1 - t_2)^2 = 6.25$ implies that the absolute difference $|t_1 - t_2|$ is 2.5. The area of the triangle $\triangle POQ$ formed by the vertex at the origin $O(0,0)$ and the points $P$ and $Q$ is calculated using the coordinates $(0,0), (t_1^2, 2t_1),$ and $(t_2^2, 2t_2)$. The area formula $\frac{1}{2} |x_1 y_2 - x_2 y_1|$ results in the expression $\frac{1}{2} |2t_1^2 t_2 - 2t_2^2 t_1|$, which simplifies to $|t_1 t_2 (t_1 - t_2)|$.

Substituting the known relationship $t_1 t_2 = -1$ and the absolute difference $|t_1 - t_2| = 2.5$ into the area expression gives $|-1 \times 2.5| = 2.5$ square units. Multiplying this area by 10 yields the final result of 25.

The geometry of the parabola $y^{2} = 2x$ dictates that the slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{1}{y}$. At the specific point $P(2, 2)$, this slope is $\frac{1}{2}$, making the equation of the tangent $x - 2y + 2 = 0$. By setting $y = 0$, we locate the intersection point $Q$ on the $x$-axis at $(-2, 0)$. Because the normal is perpendicular to the tangent at $P$, its slope must be $-2$, leading to the normal equation $y - 2 = -2(x - 2)$, which simplifies to $y = -2x + 6$.

Substituting the linear equation of the normal into the parabola's definition $x = \frac{y^{2}}{2}$ allows us to solve for the intersection point $R$. This substitution results in the quadratic equation $\frac{y^{2}}{2} = - \frac{1}{2}y + 3$, or more simply $y^{2} + y - 6 = 0$, which factors into $(y - 2)(y + 3) = 0$. Since $y = 2$ corresponds to point $P$, the other solution $y = -3$ provides the $y$-coordinate for $R$, resulting in the coordinates $R\left(\frac{9}{2}, -3\right)$.

With the vertices $P(2, 2)$, $Q(-2, 0)$, and $R\left(\frac{9}{2}, -3\right)$ established, the area of $\triangle PQR$ is found using the coordinate geometry determinant formula. Calculating $\frac{1}{2} | 2(0 - (-3)) + (-2)(-3 - 2) + \frac{9}{2}(2 - 0) |$ yields $\frac{1}{2} | 6 + 10 + 9 |$, which simplifies to exactly $\frac{25}{2}$ sq. units.

Finding the locus of a midpoint involves translating the parametric coordinates of a moving point on the original parabola into the Cartesian coordinates of the new curve. Given the parabola $y^{2}=4 a x$ with its focus at $(a, 0)$, we express any arbitrary point on the parabola as $(at^{2}, 2at)$. The midpoint $(h, k)$ of the line segment joining this point to the focus is defined by $h = \frac{at^{2}+a}{2}$ and $k = \frac{2at+0}{2}$.

By isolating the parameter $t$ from these equations, we find $t = \frac{k}{a}$ and $t^{2} = \frac{2h-a}{a}$. Substituting the expression for $t$ into the equation for $t^{2}$ yields the relationship $\frac{k^{2}}{a^{2}} = \frac{2h-a}{a}$, which simplifies to $k^{2} = a(2h-a)$ or $k^{2} = 2a(h - \frac{a}{2})$. Replacing the variables $(h, k)$ with the general coordinates $(x, y)$, we obtain the locus $y^{2} = 2a(x - \frac{a}{2})$.

This locus represents a parabola in the standard form $Y^{2} = 4AX$, where the shifted coordinate $X$ is $x - \frac{a}{2}$ and the focal length $A$ is $\frac{a}{2}$. For any parabola of the form $Y^{2} = 4AX$, the directrix is given by the equation $X = -A$. Applying this to the derived equation results in $x - \frac{a}{2} = -\frac{a}{2}$, which simplifies to $x = 0$.

To determine the path described by the midpoint, we must express the coordinates of the midpoint $(h, k)$ in terms of the original coordinates $(x_0, y_0)$ of a variable point $P$ on the parabola $y_0 = 4x_0^2 + 1$. The foot of the perpendicular $F$ drawn from $P$ to the line $y=x$ is the intersection of the line $y-y_0 = -1(x-x_0)$ and the line $x=y$, which results in the coordinates $F = (\frac{x_0+y_0}{2}, \frac{x_0+y_0}{2})$. Since $M(h, k)$ is the midpoint of $P$ and $F$, its coordinates are calculated as $h = \frac{x_0 + \frac{x_0+y_0}{2}}{2} = \frac{3x_0+y_0}{4}$ and $k = \frac{y_0 + \frac{x_0+y_0}{2}}{2} = \frac{x_0+3y_0}{4}$.

By solving this system of linear equations for $x_0$ and $y_0$, we obtain $x_0 = \frac{3h-k}{2}$ and $y_0 = \frac{3k-h}{2}$. Substituting these expressions into the original parabola equation $y_0 = 4x_0^2 + 1$ produces $\frac{3k-h}{2} = 4\left(\frac{3h-k}{2}\right)^2 + 1$. This simplifies to $\frac{3k-h}{2} = (3h-k)^2 + 1$, which further rearranges to $3k-h = 2(3h-k)^2 + 2$. By shifting terms and replacing the variables $h$ and $k$ with $x$ and $y$ to represent the locus, we define the curve as $2(3x-y)^2 + (x-3y) + 2 = 0$.

A parabola with its vertex at $V(2, 0)$ and focus at $F(4, 0)$ possesses a focal distance of $a = 2$, identifying its equation as $y^2 = 8(x - 2)$ and situating its latus rectum along the vertical line $x = 4$. Drawing tangents from the origin to this curve involves solving the quadratic equation $m^2x^2 - 8x + 16 = 0$ for tangency, which yields slopes $m = \pm 1$ and confirms that the points of tangency $S(4, 4)$ and $R(4, -4)$ fall exactly on the latus rectum. This creates a triangle $\Delta SOR$ with a base $SR$ of 8 units and a height of 4 units, corresponding to the distance of the origin from the line $x = 4$, making the final area $0.5 \times 8 \times 4 = 16$.

For a parabola defined by $y^2 = 8x$, the directrix is determined by the vertical line $x = -2$. At the point $P(2, -4)$, the tangent line is found using the formula $yy_1 = 4(x + x_1)$, which results in $-4y = 4(x + 2)$, simplifying to the linear equation $x + y + 2 = 0$. Since the slope of this tangent is $-1$, the normal line passing through $P$ must have a perpendicular slope of $1$, leading to the equation $y + 4 = 1(x - 2)$, or $x - y = 6$.

Substituting the directrix $x = -2$ into the tangent equation yields $y = 0$, identifying point $A$ as $(-2, 0)$. Similarly, substituting $x = -2$ into the normal equation results in $-8 = y$, identifying point $B$ as $(-2, -8)$. Because the figure $AQBP$ forms a square, its diagonals $AB$ and $PQ$ must bisect each other at the same midpoint. The midpoint of diagonal $AB$ is calculated as $\left( \frac{-2 - 2}{2}, \frac{0 - 8}{2} \right) = (-2, -4)$.

Equating this to the midpoint of diagonal $PQ$, which is $\left( \frac{a + 2}{2}, \frac{b - 4}{2} \right)$, provides the system $\frac{a + 2}{2} = -2$ and $\frac{b - 4}{2} = -4$. Solving these equations gives the coordinates $a = -6$ and $b = -4$. Consequently, the expression $2a + b$ is evaluated as $2(-6) - 4 = -16$.

The length of the latus rectum of any parabola is geometrically defined as four times the focal length, which represents the distance between its vertex and its focus. Since the vertex lies at $R$ and the focus lies at $S$ along the positive X-axis, the focal length is simply the difference between these two points, calculated as $S - R$. Multiplying this distance by four gives the final length of the latus rectum as $4(S - R)$.

The parabola $y^{2} = -64x$ has its focus at $(-16, 0)$. Since the focal chord $y = mx + c$ must pass through this focus, substituting the coordinates into the line equation gives $0 = -16m + c$, which establishes that $c = 16m$. This relation anchors the line's properties to the geometry of the parabola.

The line is also tangent to the circle $(x+10)^{2} + y^{2} = 4$, which has its center at $(-10, 0)$ and a radius of $2$. For a line to be tangent to a circle, the perpendicular distance from the circle's center to the line must equal the radius. Applying the distance formula to the line $mx - y + c = 0$ gives $\frac{|m(-10) - 0 + c|}{\sqrt{m^{2} + 1}} = 2$. Substituting $c = 16m$ into this expression yields $|16m - 10m| = 2\sqrt{m^{2} + 1}$, which simplifies to $6m = 2\sqrt{m^{2} + 1}$ since $m > 0$.

Squaring both sides results in $36m^{2} = 4(m^{2} + 1)$, which reduces to $9m^{2} = m^{2} + 1$, or $8m^{2} = 1$. Solving for the slope gives $m = \frac{1}{2\sqrt{2}}$, which leads to the intercept $c = 16 \left( \frac{1}{2\sqrt{2}} \right) = \frac{8}{\sqrt{2}}$. Calculating the final expression $4\sqrt{2}(m+c)$ involves substituting these values to get $4\sqrt{2} \left( \frac{1}{2\sqrt{2}} + \frac{8}{\sqrt{2}} \right)$, which simplifies to $4\sqrt{2} \left( \frac{17}{2\sqrt{2}} \right)$, ultimately resulting in 34.

The shortest distance between a parabola and a line occurs at the specific point on the curve where the tangent line is parallel to the given line. We begin by identifying the slope of the line $x-y=1$, which is 1, as the line can be rearranged into the form $y=x-1$. By differentiating the equation of the parabola $x^2=2y$, we obtain $2x=2\frac{dy}{dx}$, which simplifies to $\frac{dy}{dx}=x$ for the slope of the tangent at any point on the curve.

Setting the slope of the tangent equal to the slope of the line requires $x=1$. Substituting this value into the parabola's equation $1^2=2y$ yields $y=\frac{1}{2}$, which identifies the closest point on the curve as $(1, \frac{1}{2})$. We find the perpendicular distance from this point to the line $x-y-1=0$ by applying the distance formula:
$d = \frac{|1(1) + (-1)(\frac{1}{2}) - 1|}{\sqrt{1^2 + (-1)^2}}$
Simplifying the numerator results in the absolute value of $-\frac{1}{2}$, which is $\frac{1}{2}$. Dividing this result by the denominator $\sqrt{2}$ yields the shortest distance of $\frac{1}{2\sqrt{2}}$.

Reflection across the line $y=x$ simply swaps the $x$ and $y$ coordinates of every point, which effectively transforms the original parabola $y^2 = 4x$ into its reflected counterpart, $x^2 = 4y$. This resulting locus $C$ is the mirrored curve where the relationship between the variables is perfectly inverted.

Determining the tangent at $P(2,1)$ involves differentiating the locus equation $4y = x^2$ with respect to $x$, which yields $4 \frac{dy}{dx} = 2x$, or simplified, $\frac{dy}{dx} = \frac{x}{2}$. At the coordinate $x = 2$, the slope of the tangent line is calculated as $\frac{2}{2} = 1$. Substituting this slope and the coordinates of point $P$ into the linear point-slope form $y - 1 = 1(x - 2)$ results in the equation $x - y = 1$.

The geometric 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 $c^2 = a^2m^2 + B^2$. We can identify the necessary constants $m$ and $c$ by finding the tangent line to the parabola $y^2 = 4x - 20$ at the point $(6, 2)$. Differentiating the parabola's equation with respect to $x$ gives $2y \frac{dy}{dx} = 4$, which simplifies to $\frac{dy}{dx} = \frac{2}{y}$. Evaluating this derivative at $(6, 2)$ provides a slope $m = \frac{2}{2} = 1$. Using the point-slope form, the tangent line equation is $y - 2 = 1(x - 6)$, which simplifies to $y = x - 4$.

With the tangent line established as $y = 1x - 4$, we identify the slope $m = 1$ and the intercept $c = -4$. Substituting these values into the tangency condition for the ellipse $\frac{x^2}{2} + \frac{y^2}{b} = 1$, where $a^2 = 2$ and $B^2 = b$, results in $(-4)^2 = 2(1)^2 + b$. This simplifies to $16 = 2 + b$, which leads to the final value $b = 14$.

Any pair of tangents to a parabola that intersect at a right angle will always meet on the parabola's directrix. For the given equation $y^{2}=16(x-3)$, we identify this as a horizontal parabola in the form $(y-k)^{2}=4a(x-h)$, where the vertex shift is $h=3$ and the focal parameter $4a=16$ indicates $a=4$. The directrix of such a parabola is located at a distance $a$ behind the vertex, following the standard relationship $x-h = -a$. Substituting our specific values into this expression results in:

$x - 3 = -4$

Rearranging the terms of this equation leads to $x+1=0$.

The point on a curve closest to a given line occurs where the tangent to the curve is parallel to that line, meaning they must share the same slope. By differentiating the equation of the parabola $y = x^2 + 4$ with respect to $x$, we obtain the slope function $\frac{dy}{dx} = 2x$. We equate this to the slope of the line $y = 4x - 1$, which is $4$, leading to the equation $2x = 4$. Solving for $x$ gives $x = 2$, and substituting this value back into the original parabola equation results in $y = 2^2 + 4 = 8$, which determines the coordinates of point $P$ to be $(2, 8)$.

A line tangent to both a parabola and a circle requires the perpendicular distance from the circle's center to the line to equal its radius. The tangent to $y^{2}=8x$ at $(2,-4)$ is obtained via $yy_{1}=4(x+x_{1})$, which results in $-4y=4(x+2)$ and simplifies to $x+y+2=0$. Since the circle $x^{2}+y^{2}=a$ has a center at $(0,0)$ and a radius of $\sqrt{a}$, the perpendicular distance from the origin to the line $x+y+2=0$ is calculated as $\frac{|0+0+2|}{\sqrt{1^{2}+1^{2}}}=\sqrt{2}$. Equating this perpendicular distance to the radius $\sqrt{a}$ confirms that $\sqrt{a}=\sqrt{2}$, which means $a=2$.

The parabola features a vertex at $\left(\frac{1}{2}, \frac{3}{4}\right)$ and a horizontal directrix $y=\frac{1}{2}$, placing it in the form $(x-h)^2 = 4a(y-k)$. Since the vertical distance between the vertex and the directrix is $\frac{1}{4}$, the focal parameter $4a$ equals $1$, resulting in the equation $(x-\frac{1}{2})^2 = y-\frac{3}{4}$. Locating the point $P$ where the line $x=-\frac{1}{2}$ intersects this curve requires substituting $x=-\frac{1}{2}$ into the equation, yielding $(-1)^2 = y-\frac{3}{4}$, which simplifies to $y=\frac{7}{4}$. Thus, the coordinates of $P$ are $\left(-\frac{1}{2}, \frac{7}{4}\right)$.

Determining the normal at $P$ involves finding the derivative of the parabola equation, $2(x-\frac{1}{2}) = \frac{dy}{dx}$, which evaluates to $-2$ at $x=-\frac{1}{2}$. Because the normal is perpendicular to the tangent, its slope is $\frac{1}{2}$, leading to the equation $y-\frac{7}{4} = \frac{1}{2}\left(x+\frac{1}{2}\right)$, which rearranges to $x=2y-4$. Substituting this linear relation back into the parabola equation creates the quadratic expression $(2y-4-\frac{1}{2})^2 = y-\frac{3}{4}$, which simplifies to $4y^2-19y+21=0$. Factoring this quadratic reveals roots at $y=\frac{7}{4}$ (corresponding to point $P$) and $y=3$ (corresponding to point $Q$). With $y=3$, the $x$-coordinate is determined to be $2$, identifying $Q$ as $(2,3)$. Calculating the squared distance between $P\left(-\frac{1}{2}, \frac{7}{4}\right)$ and $Q(2,3)$ results in $\left(2+\frac{1}{2}\right)^2 + \left(3-\frac{7}{4}\right)^2$, which equals $\left(\frac{5}{2}\right)^2 + \left(\frac{5}{4}\right)^2 = \frac{25}{4} + \frac{25}{16} = \frac{100}{16} + \frac{25}{16} = \frac{125}{16}$.

The condition for a line to be tangent to the parabola $y^2=4x$ follows the slope-intercept form $y = mx + \frac{1}{m}$, where $m$ is the slope. Rearranging this into the general form $m^2x - my + 1 = 0$, we impose the condition that the perpendicular distance from the center of the circle at $(3,0)$ to this line must equal the radius of 3. Setting the distance equal to 3 results in:
$\frac{|3m^2 + 1|}{\sqrt{m^4 + m^2}} = 3$

Squaring both sides leads to $(3m^2+1)^2 = 9(m^4+m^2)$, which simplifies to $9m^4 + 6m^2 + 1 = 9m^4 + 9m^2$, or $3m^2 = 1$. Since the points of contact are in the first quadrant, we take the positive slope $m = \frac{1}{\sqrt{3}}$. Substituting this value into the parametric point of contact formula $(\frac{1}{m^2}, \frac{2}{m})$ for the parabola yields the point $(c, d) = (3, 2\sqrt{3})$.

Substituting the tangent line equation $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$ into the circle's equation $(x-3)^2 + y^2 = 9$ gives $(x-3)^2 + (\frac{x}{\sqrt{3}} + \sqrt{3})^2 = 9$. Expanding this leads to $x^2 - 6x + 9 + \frac{x^2}{3} + 2x + 3 = 9$, which simplifies to the quadratic $\frac{4}{3}x^2 - 4x + 3 = 0$. Factoring this expression as $(2x-3)^2 = 0$ identifies the x-coordinate $a = \frac{3}{2}$. Calculating $2(a+c)$ then results in $2(\frac{3}{2} + 3)$, which equals 9.

The line $2x + y = 1$ has a slope of $-2$, implying that any line perpendicular to it must have a slope of $m = 1/2$. For the parabola $y^2 = 6x$, where the standard form $y^2 = 4ax$ gives $a = 3/2$, the equation of the tangent line is found using $y = mx + a/m$. Substituting the values of $m$ and $a$ results in $y = \frac{1}{2}x + \frac{3/2}{1/2}$, which simplifies to $y = \frac{1}{2}x + 3$ or:
$x - 2y + 6 = 0$
Evaluating the given points, we find that $(-6, 0)$, $(4, 5)$, and $(0, 3)$ all satisfy this equation, as they produce a result of zero. However, substituting $x = 5$ and $y = 4$ into the equation yields $5 - 2(4) + 6 = 3$, confirming that this point does not lie on the tangent.

The existence of three distinct normals to the parabola $y^{2}=2x$ passing through a point $(a, 0)$ relies on finding the roots of the cubic equation that determines the slopes of these normals. By using the standard form $y^{2}=4Ax$, where $A=0.5$, the general equation of a normal is $y=-tx+t+0.5t^{3}$. When this normal passes through the point $(a, 0)$, substituting the coordinates results in the cubic equation $0.5t^{3}+(1-a)t=0$, which simplifies to $t(t^{2}+2-2a)=0$.

This factorization shows that one normal is always the x-axis, corresponding to the root $t=0$. For the other two distinct real normals to exist, the quadratic component $t^{2}=2a-2$ must have a positive value, requiring that $2a-2>0$. Solving this inequality directly confirms that $a$ must be greater than $1$.

The shortest distance between a point and a parabola is defined by the normal line passing through that point on the curve. For the parabola $y^2 = 6x$, which is in the standard form $y^2 = 4ax$, we identify $4a = 6$, giving $a = \frac{3}{2}$. We can represent any point on this curve using the parameter $t$ as $(\frac{3}{2}t^2, 3t)$.

The equation for the normal at any point $t$ on this parabola is given by $tx + y = 2at + at^3$, which becomes $tx + y = 3t + \frac{3}{2}t^3$. Since this line must contain the point $(3, \frac{3}{2})$, substituting these coordinates yields $t(3) + \frac{3}{2} = 3t + \frac{3}{2}t^3$. By subtracting $3t$ from both sides, we simplify the equation to $\frac{3}{2} = \frac{3}{2}t^3$, which implies $t^3 = 1$ and therefore $t = 1$. Plugging this value back into the parametric coordinates identifies the nearest point $(\alpha, \beta)$ as $(\frac{3}{2}, 3)$. The final required calculation is $2(\alpha + \beta) = 2(\frac{3}{2} + 3) = 2(\frac{9}{2}) = 9$.