Parabola – JEE Main MathematicsPractice Questions & PYQs

Any line tangent to the parabola $y^2 = 4x$ can be expressed in the form $y = mx + \frac{1}{m}$, which rearranges to the linear equation $m^2x - my + 1 = 0$. For this same line to serve as a tangent to the circle $x^2 + y^2 - 6x = 0$, the perpendicular distance from the center of the circle to the line must be exactly equal to the circle's radius. Rewriting the circle's equation as $(x-3)^2 + y^2 = 3^2$ reveals a center at $(3, 0)$ and a radius of $3$. Applying the perpendicular distance formula $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} = r$ to the tangent line and the center $(3, 0)$ gives the relation $\frac{|3m^2 + 1|}{\sqrt{m^4 + m^2}} = 3$. Squaring both sides of this equality leads to the equation $9m^4 + 6m^2 + 1 = 9m^4 + 9m^2$, which simplifies directly to $3m^2 = 1$, or $m = \pm \frac{1}{\sqrt{3}}$. Substituting the slope $m = \frac{1}{\sqrt{3}}$ into the initial tangent form $y = mx + \frac{1}{m}$ yields $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$, which is equivalent to the expression $\sqrt{3}y = x + 3$.

To maximize the area of a triangle formed by a fixed base $PQ$ and a variable point $X$ on a parabola, the tangent to the curve at $X$ must be parallel to the chord $PQ$. The slope of the chord connecting $P(4, -4)$ and $Q(9, 6)$ is calculated as $\frac{6 - (-4)}{9 - 4} = 2$. Differentiating the parabola equation $y^2 = 4x$ with respect to $x$ yields $2y \frac{dy}{dx} = 4$, which simplifies to the derivative $\frac{dy}{dx} = \frac{2}{y}$. Setting this derivative equal to the slope of the chord, $\frac{2}{y} = 2$, identifies the ordinate of $X$ as $y = 1$, which confirms the abscissa $x = \frac{1}{4}$ via the parabola equation.

With the coordinates of $X(\frac{1}{4}, 1)$ established, the area of triangle $PXQ$ is calculated using the formula $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Substituting the points $P(4, -4)$, $Q(9, 6)$, and $X(\frac{1}{4}, 1)$ into this expression results in $\frac{1}{2} |4(6 - 1) + 9(1 - (-4)) + \frac{1}{4}(-4 - 6)|$, which simplifies to $\frac{1}{2} |20 + 45 - 2.5| = \frac{125}{4}$.

The shortest distance between a curve and a line is always found at the point on the curve where the tangent is parallel to the given line. Because the given line $y = x$ has a slope of $1$, we look for a tangent to the parabola $y^2 = x - 2$ that shares this same slope. Using the geometric properties of this parabola, the equation for this parallel tangent is $y = x - 2 + \frac{1}{4}$, which simplifies to $y = x - \frac{7}{4}$. By treating these two expressions as the parallel lines $x - y = 0$ and $x - y - \frac{7}{4} = 0$, the perpendicular distance between them is calculated as $\frac{|-7/4|}{\sqrt{1^2 + (-1)^2}}$, yielding a final result of $\frac{7}{4\sqrt{2}}$.

Identifying the specific point of intersection between the parabola $y^2 = 4x$ and the circle $x^2 + y^2 = 5$ allows us to define the equation of the tangent line necessary to evaluate the given coordinates. Substituting $y^2 = 4x$ into the circle equation yields $x^2 + 4x - 5 = 0$, which factors into $(x + 5)(x - 1) = 0$, providing the valid intersection $x = 1$ in the first quadrant, corresponding to $y = 2$. With the point of tangency established as $(1, 2)$, the standard tangent formula $y y_1 = 2(x + x_1)$ becomes $2y = 2(x + 1)$, which simplifies to the linear equation $x - y + 1 = 0$. The point $\left(\frac{3}{4}, \frac{7}{4}\right)$ is the only option that satisfies this equation, as $\frac{3}{4} - \frac{7}{4} + 1 = 0$.

A line $y=mx+c$ is tangent to the circle $x^2+y^2=r^2$ if the squared intercept satisfies $c^2=r^2(1+m^2)$, and it is tangent to the parabola $y^2=4Ax$ when the condition $c=A/m$ is met. Rearranging the line equation $ax+y=c$ into the slope-intercept form $y=-ax+c$ reveals that the slope $m$ is $-a$. For this line to touch the circle $x^2+y^2=1$, we apply the condition $c^2=1(1+(-a)^2)$, which simplifies to $c^2=1+a^2$.

Simultaneously, for the line to touch the parabola $y^2=4\sqrt{2}x$, we identify $A=\sqrt{2}$ and use the tangency condition $c = A/m$, yielding $c = -\sqrt{2}/a$. Squaring this result gives $c^2=2/a^2$, which implies that $a^2=2/c^2$. Substituting this expression into the condition derived from the circle leads to the equation $c^2=1+2/c^2$, which simplifies to the biquadratic form $c^4-c^2-2=0$. Factoring this expression as $(c^2-2)(c^2+1)=0$ and noting that $c^2$ must be a positive real value, we find $c^2=2$, which means the magnitude $|c|$ is equal to $\sqrt{2}$.

A common normal between two parabolas exists if there is a slope $m$ that satisfies the equation of a normal for both curves simultaneously. The equation of a normal to a parabola of the form $y^2 = 4Ax$ is $y = mx - 2Am - Am^3$. For the first parabola $y^2 = 4b(x - c)$, the vertex is shifted to $(c, 0)$, so the normal equation becomes $y = m(x - c) - 2bm - bm^3$. For the second parabola $y^2 = 8ax$, which follows the standard form with $A = 2a$, the normal equation is $y = mx - 4am - 2am^3$.

Equating these two expressions allows us to find the intersection of the normal lines, leading to $(c + 2b)m + bm^3 = 4am + 2am^3$, which simplifies to $m(4a - c - 2b) = m^3(b - 2a)$. This equation is satisfied when $m = 0$, representing the x-axis. Because the axis of symmetry is always a normal line for any parabola, and both given equations describe parabolas whose axes of symmetry lie on the x-axis, the x-axis serves as a shared common normal regardless of the specific values assigned to the parameters. Since this condition is satisfied for any triad of $(a, b, c)$, every provided option represents a valid configuration.

For any parabola defined by $y^2 = 4ax$, any point on the curve can be expressed parametrically as $(at^2, 2at)$. Comparing the given equation $y^2 = 16x$ to the standard form reveals that $a = 4$. By substituting the coordinates $(1, 4)$ into the parametric form, we solve $2(4)t = 4$ to find the parameter $t_1 = \frac{1}{2}$ for the first endpoint. Because the line segment is a focal chord, the parameters of the two endpoints must satisfy the relation $t_1 t_2 = -1$, which leads to $t_2 = -2$. Determining the coordinates of the second endpoint $Q$ gives $(4(-2)^2, 2(4)(-2))$, resulting in $(16, -16)$. Calculating the length of the chord using the distance formula between $(1, 4)$ and $(16, -16)$ yields $\sqrt{(16-1)^2 + (-16-4)^2}$, which simplifies to $\sqrt{15^2 + (-20)^2} = \sqrt{225 + 400} = \sqrt{625} = 25$.

With the vertex located at $(2, 0)$ and the focus at $(4, 0)$, the focal length $a$ is $4 - 2 = 2$, which defines the parabola by the equation $y^2 = 8(x - 2)$. Any point lying on this curve must satisfy this equality. While the other points fulfill this requirement, testing the point $(8, 6)$ produces $6^2 = 36$ on the left side and $8(8 - 2) = 48$ on the right, confirming it does not lie on the parabola.

A line tangent to the parabola $y^{2}=16x$ can be expressed in terms of its slope $m$ as $y=mx+\frac{4}{m}$, where the parameter $a$ is identified as $4$. For this line to also serve as a tangent to the hyperbola $xy=-4$, the intersection of the line and the hyperbola must occur at exactly one point, meaning the discriminant of the resulting quadratic equation must be zero. Substituting the expression for $y$ into the hyperbola equation gives $x(mx+\frac{4}{m})=-4$, which expands to $mx^2+\frac{4x}{m}+4=0$. Multiplying the entire equation by $m$ yields $m^2x^2+4x+4m=0$, and setting the discriminant $b^{2}-4ac$ equal to zero results in $16-16m^3=0$. Solving $m^{3}=1$ gives a slope of $m=1$, which, when substituted back into the tangent formula, results in $y=x+4$ or $x-y+4=0$.

The geometry of a parabola dictates that the tangent and normal at a given point have specific relationships with the parabola's axis. For the parabola $y^2 = 16x$ at point $P(16, 16)$, the tangent meets the x-axis at $A(-16, 0)$, while the normal intersects at $B(24, 0)$. Since the circle passes through $P, A,$ and $B$, its center $C$ must lie on the perpendicular bisector of the chord $AB$, which is the vertical line $x = \frac{-16+24}{2} = 4$. Given that $C$ is equidistant from $A$ and $B$, its vertical coordinate is 0, establishing the center at $C(4, 0)$.

To find $\tan\theta$ for $\angle CPB$, we calculate the slopes of lines $CP$ and $BP$. The slope $m_1$ of $CP$ is $\frac{16-0}{16-4} = \frac{4}{3}$, and the slope $m_2$ of $BP$ is $\frac{16-0}{16-24} = -2$. Using the formula $\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$, we substitute these values to obtain $\tan\theta = \left|\frac{4/3 - (-2)}{1 + (4/3)(-2)}\right|$. This expression simplifies to $\left|\frac{10/3}{-5/3}\right|$, which yields 2.

The condition for a line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$ is $c = a/m$, and to the parabola $x^2 = 4ay$ is $c = -am^2$. Given the length of the latus rectum is 3, $4a = 3$ results in $a = 3/4$, defining the parabolas as $y^2 = 3x$ and $x^2 = 3y$. Equating the values of $c$ for a common tangent gives $\frac{3/4}{m} = -\frac{3}{4}m^2$, which simplifies to $1/m = -m^2$ or $m^3 = -1$, yielding $m = -1$. Substituting $m = -1$ into the tangency condition $c = \frac{3/4}{-1}$ gives $c = -3/4$, which results in the line $y = -x - 3/4$, and multiplying by 4 produces $4x + 4y + 3 = 0$, or $4(x+y)+3=0$.

For a parabola defined by $y^2 = 4ax$, the tangent at point $t$ follows the equation $ty = x + at^2$. Comparing $y^2 = 8x$ to the standard form identifies $a = 2$. Substituting the external point $(-8, 0)$ into the tangent equation gives $0 = -8 + 2t^2$, which simplifies to $t^2 = 4$, yielding parameter values $t = 2$ and $t = -2$.

The points of contact $P$ and $Q$ are $(at^2, 2at)$, resulting in $P = (2(2^2), 2(2)(2)) = (8, 8)$ and $Q = (2(-2)^2, 2(2)(-2)) = (8, -8)$. With the focus $F$ located at $(2, 0)$, triangle $PFQ$ has a vertical base $PQ$ with length $8 - (-8) = 16$ and an altitude from $F$ to the line $x=8$ equal to $8 - 2 = 6$. The area is $\frac{1}{2} \times 16 \times 6 = 48$ square units.

The shortest distance between a point and a curve is measured along the normal passing through that point. Rewriting the circle equation $x^2+y^2+6x+8=0$ as $(x+3)^2+y^2=1$ identifies the center as $C(-3, 0)$. Any point $P$ on the parabola $x^2=4y$ can be expressed parametrically as $(2t, t^2)$, and the equation of the normal at this point is given by $x+ty=2t+t^3$. Substituting the coordinates of the center $C(-3, 0)$ into this normal equation yields $-3+t(0) = 2t+t^3$, which simplifies to the cubic equation $t^3+2t+3=0$. By observation, $t=-1$ satisfies this equation, which determines the point $P$ as $(-2, 1)$. The tangent line to the parabola at a point $(x_1, y_1)$ is given by $x x_1 = 2(y+y_1)$, and substituting $x_1=-2$ and $y_1=1$ results in $-2x = 2(y+1)$, which simplifies to $x+y+1=0$.

The focal distance of a point $(x_1, y_1)$ on the parabola $y^2=4ax$ is given by $x_1+a$. For the parabola $y^2=8x$, we identify $a=2$, and setting the focal distance $x_1+2=8$ yields $x_1=6$, which leads to $y_1^2=48$ and $y_1 = \pm 4\sqrt{3}$.

The slope of the tangent at $(x_1, y_1)$ is $\frac{dy}{dx} = \frac{4}{y_1}$, so the slope of the normal $m$ is $-\frac{y_1}{4} = \mp \sqrt{3}$. Given the standard normal equation $y=mx-2am-am^3$, we substitute $a=2$ and $m = \pm \sqrt{3}$ to calculate $c = -4m - 2m^3 = -4(\pm \sqrt{3}) - 2(\pm \sqrt{3})^3 = \mp 4\sqrt{3} \mp 6\sqrt{3} = \mp 10\sqrt{3}$, resulting in $|c|=10\sqrt{3}$.

Common tangents to the parabola $x^2=4y$ and the circle $x^2+y^2=4$ are symmetric with respect to the y-axis, which implies that their intersection point $P$ must lie on the y-axis at some coordinate $(0, c)$. A line with the equation $y = mx + c$ is tangent to the circle $x^2+y^2=4$ if it satisfies the condition $c^2 = 4(1+m^2)$, which represents the perpendicular distance from the origin to the line being equal to the radius of the circle.

For this same line to be tangent to the parabola $x^2=4y$, we substitute $y = mx + c$ into the parabola's equation to obtain $x^2 = 4(mx + c)$, or $x^2 - 4mx - 4c = 0$. For this quadratic equation to have exactly one solution, its discriminant must be zero, giving $(-4m)^2 - 4(1)(-4c) = 0$, which simplifies to $16m^2 + 16c = 0$, or $c = -m^2$. By substituting this expression for $c$ into our circle tangency condition, we arrive at $(-m^2)^2 = 4(1+m^2)$, which simplifies to the biquadratic equation $m^4 - 4m^2 - 4 = 0$.

Applying the quadratic formula to solve for $m^2$, we find $m^2 = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2}$, which yields $m^2 = 2 \pm 2\sqrt{2}$. Since $m^2$ must be a positive real number, we select $m^2 = 2 + 2\sqrt{2}$. Given that the y-intercept is $c = -m^2$, the coordinates of the intersection point $P$ are $(0, -(2+2\sqrt{2}))$. The distance of this point from the origin is simply the absolute value of its y-coordinate, resulting in $2+2\sqrt{2}$, which factors to $2(\sqrt{2}+1)$.

The normal to the parabola $y^2=4x$ at a point with parameter $t$ follows the linear equation $y=-tx+2t+t^3$. Since the point $Q$ with parameter $t_1$ lies on this normal, substituting its coordinates $x=t_1^2$ and $y=2t_1$ into the equation establishes the relationship $2t_1=-t(t_1^2)+2t+t^3$. Dividing the rearranged terms by $(t_1-t)$, which is non-zero because the points are distinct, reveals the relationship $t_1=-t-\frac{2}{t}$.

Squaring this relationship leads to the expression $t_1^2=t^2+\frac{4}{t^2}+4$. By applying the arithmetic mean-geometric mean inequality, the sum of the variable components $t^2+\frac{4}{t^2}$ must be at least $2\sqrt{t^2 \cdot \frac{4}{t^2}}$, which simplifies to 4. Adding the constant 4 to this minimum value produces a total minimum of 8 for $t_1^2$.

The shortest distance between a fixed point and a curve is always measured along the normal to the curve at the point of closest approach. For the parabola $y^2 = 8x$, which has the standard form $y^2 = 4ax$ with $a = 2$, any point $P$ can be expressed in parametric coordinates as $(2t^2, 4t)$. The normal to the parabola at this point is defined by the equation $y = -tx + 2t^3 + 4t$. Since this normal line must pass through the center of the given circle, $C(0, -6)$, substituting these coordinates into the normal equation yields $-6 = 2t^3 + 4t$, or $t^3 + 2t + 3 = 0$. By inspection, the value $t = -1$ satisfies this cubic equation.

Evaluating the parametric coordinates at $t = -1$ identifies the point $P$ as $(2, -4)$. The circle to be constructed is centered at $P(2, -4)$ and passes through $C(0, -6)$, meaning its squared radius is equivalent to the squared distance between these two points: $(2 - 0)^2 + (-4 - (-6))^2$, which simplifies to $2^2 + 2^2 = 8$. Using the standard center-radius form $(x - 2)^2 + (y + 4)^2 = 8$ and expanding the terms results in $x^2 - 4x + 4 + y^2 + 8y + 16 = 8$, which simplifies to the final equation $x^2 + y^2 - 4x + 8y + 12 = 0$.

The locus of a point dividing a line segment in a fixed ratio is determined by expressing the coordinates of the moving point in terms of the fixed point's coordinates and then substituting these into the parent equation. Considering the parabola $x^2 = 8y$ with vertex $O$ at the origin $(0, 0)$, any point $Q(x_1, y_1)$ on this curve satisfies the relationship $x_1^2 = 8y_1$. When a point $P(x, y)$ divides the line segment $OQ$ internally in the ratio $1 : 3$, the section formula provides $x = \frac{x_1}{4}$ and $y = \frac{y_1}{4}$, which implies $x_1 = 4x$ and $y_1 = 4y$. Substituting these expressions back into the original parabolic equation gives $(4x)^2 = 8(4y)$, which simplifies to $16x^2 = 32y$ and results in the final locus $x^2 = 2y$.

When a straight line acts as a tangent to two distinct parabolas, its slope and y-intercept must satisfy the tangency requirements for both curves simultaneously. The general equation of a tangent line with slope $m$ for the parabola $y^2 = 4x$ is given by $y = mx + \frac{1}{m}$. For the second parabola, $x^2 = -32y$, we can rewrite the equation in the standard form $x^2 = 4Ay$ as $x^2 = 4(-8)y$, where $A = -8$. The condition for a line $y = mx + c$ to be tangent to this parabola is $c = -Am^2$, which results in $c = -(-8)m^2$ or simply $c = 8m^2$.

Since the line is common to both parabolas, the y-intercepts must be identical, allowing us to set the two expressions for $c$ equal as $\frac{1}{m} = 8m^2$. Multiplying both sides by $m$ yields $8m^3 = 1$, which simplifies to $m^3 = \frac{1}{8}$. Taking the cube root of both sides gives the value for the slope as $m = \frac{1}{2}$.

When two tangents are drawn to a parabola such that they meet at a right angle, their intersection point necessarily falls on the directrix of that parabola. For a standard parabola given by $y^2 = 4ax$, the directrix is represented by the line $x = -a$. Since the given equation $y^2 = 4x$ implies that $a=1$, the directrix is defined by the equation $x = -1$. This means the locus of all such points $P$ is the vertical line $x = -1$.