Parabola – JEE Main MathematicsPractice Questions & PYQs

In a parabola, the perpendicular distance from the focus to the tangent at the vertex is equivalent to the focal length, and the length of the latus rectum is exactly four times this distance. Given the focus is $(a, a)$ and the tangent at the vertex is $x + y - a = 0$, the distance $d$ is determined using the standard point-to-line distance formula, yielding $d = \frac{|a + a - a|}{\sqrt{1^2 + 1^2}}$, which simplifies to $\frac{|a|}{\sqrt{2}}$.

With the length of the latus rectum specified as $16$, the relationship $4 \times \frac{|a|}{\sqrt{2}} = 16$ must hold true. Dividing both sides by $4$ results in $\frac{|a|}{\sqrt{2}} = 4$, which leads to $|a| = 4\sqrt{2}$.

The geometry of a parabola establishes that the perpendicular distance from its vertex to its directrix is equal to the focal parameter $a$, and the total length of the latus rectum is defined as $4a$. To find this distance for the given vertex $(2, -1)$ and directrix line $4x - 3y - 21 = 0$, we apply the standard perpendicular distance formula $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Substituting the coordinates of the vertex and the coefficients of the directrix line leads to $a = \frac{|4(2) - 3(-1) - 21|}{\sqrt{4^2 + (-3)^2}}$, which simplifies to $\frac{|8 + 3 - 21|}{5} = 2$. Since the length of the latus rectum is four times the value of $a$, the resulting length is 8.

The geometric condition that the $y$-axis bisects the segment $P Q$ implies that the average of the $x$-coordinates of point $P(x, y)$ and the $x$-intercept $Q$ is zero. The tangent at $P$ is given by $Y - y = \frac{dy}{dx}(X - x)$, and setting $Y = 0$ reveals the $x$-coordinate of $Q$ to be $x - y \frac{dx}{dy}$. Setting the midpoint of the $x$-coordinates, $\frac{x + (x - y \frac{dx}{dy})}{2}$, to zero results in the differential equation:

$2x - y \frac{dx}{dy} = 0$

Separating variables leads to $\frac{dx}{x} = \frac{2 dy}{y}$, and integrating both sides yields $\ln|x| = 2 \ln|y| + k$, which simplifies to the form $x = cy^2$ or $y^2 = \frac{1}{c}x$. Given the curve passes through $(3, 3)$, we have $3 = c(3^2)$, which gives $c = \frac{1}{3}$, resulting in the parabola $y^2 = 3x$. Comparing this to the standard form $y^2 = 4ax$, the length of the latus rectum is the coefficient of $x$, which is 3.

The axis of the parabola $P_1$ passes through the vertex $(3,2)$ and the focus $(4,4)$, which means the directrix must be a line perpendicular to this axis. Since the vertex is the midpoint between the focus and the intersection point $A$ of the directrix and the axis, the point $A$ must be $(2,0)$. Because the axis has a slope of 2, the directrix must have a slope of $-\frac{1}{2}$ and pass through $(2,0)$, leading to the equation $x+2y=2$.

Reflecting this directrix across the line $x+2y=6$ results in the directrix for $P_2$. Since the original directrix and the mirror line are parallel, the reflected line will also be parallel and take the form $x+2y=k$. We find $k$ by reflecting point $A(2,0)$ across $x+2y=6$ to obtain a point $B(x_2, y_2)$ that lies on the new directrix. Using the reflection formula, the coordinates of $B$ are determined by $\frac{x_2-2}{1} = \frac{y_2-0}{2} = -2 \left( \frac{2+0-6}{1^2+2^2} \right)$, which simplifies to $\frac{8}{5}$. This gives $B = \left( \frac{18}{5}, \frac{16}{5} \right)$. Substituting these coordinates into the form $x+2y=k$ yields $\frac{18}{5} + 2\left(\frac{16}{5}\right) = \frac{50}{5} = 10$, so the final equation is $x+2y=10$.

The conic defined by the parametric equations $x=2t$ and $y=\frac{t^2}{3}$ can be expressed in Cartesian coordinates by substituting $t = \frac{x}{2}$, yielding $y = \frac{x^2}{12}$, or $x^2 = 12y$. This is a parabola with the standard form $x^2 = 4ay$, which identifies the focus $S$ at the coordinates $(0,3)$. As $t$ approaches $1$, the point $A$ on the conic is positioned at $(2, 1/3)$. We designate point $B$ on the $y$-axis of the parabola as $(0, \beta)$.

The condition that the segments $SA$ and $BA$ are perpendicular requires that the product of their slopes equals $-1$. The slope of $SA$ is calculated as $\frac{1/3 - 3}{2 - 0} = \frac{-8/3}{2} = -\frac{4}{3}$. With the slope of $BA$ being $\frac{1/3 - \beta}{2 - 0}$, the perpendicularity condition gives $\left(-\frac{4}{3}\right) \left(\frac{1/3 - \beta}{2}\right) = -1$. Simplifying this expression, we have $\left(-\frac{2}{3}\right) \left(\frac{1}{3} - \beta\right) = -1$, which means $\frac{1}{3} - \beta = \frac{3}{2}$, resulting in $\beta = -\frac{7}{6}$.

The ordinate $k$ of the centroid of triangle $SAB$ is the average of the $y$-coordinates of its vertices: $S(0,3)$, $A(2, 1/3)$, and $B(0, -7/6)$. Therefore, $k = \frac{3 + 1/3 - 7/6}{3}$. Converting these values to a common denominator of $6$, the numerator becomes $\frac{18 + 2 - 7}{6} = \frac{13}{6}$. Dividing this by $3$ gives the final value $k = \frac{13}{18}$. Since the points are fixed at $t=1$, the limit of $k$ as $t$ approaches $1$ is $\frac{13}{18}$.

The center of the circle $x^2+y^2-10 x-14 y+65=0$ is located at $(5, 7)$, which represents the point that the tangent line to the parabola $y^2=8 x$ must intersect. For a parabola of this form, the equation of the tangent at a parameter $t$ is $yt=x+2t^2$, and since this line must pass through $(5, 7)$, we can write $7t=5+2t^2$, which simplifies to the quadratic equation $2t^2-7t+5=0$.

Solving this quadratic provides two valid parameter values, $t=1$ and $t=\frac{5}{2}$, which correspond to the two points of tangency $P(a, b)$ on the parabola where $a=2t^2$ and $b=4t$. For $t=1$, the coordinates are $a=2$ and $b=4$, while for $t=\frac{5}{2}$, the coordinates are $a=12.5$ and $b=10$. Calculating the product of all possible $a$ values yields $A=2 \times 12.5=25$, and the product of all possible $b$ values yields $B=4 \times 10=40$, resulting in a final sum of $A+B=65$.

The area of a quadrilateral with two vertices on the same coordinate axis is efficiently computed by summing the areas of the two triangles formed by those vertices and the remaining points. For the quadrilateral $OMFN$ with vertices $O(0,0)$ and $F(2\sqrt{2},0)$ on the $x$-axis, the line segment $OF$ serves as a common base of length $2\sqrt{2}$. The area is the sum of the areas of $\triangle OFM$ and $\triangle OFN$, which simplifies to $\frac{1}{2} \cdot |OF| \cdot (|y_1| + |y_2|)$, where $y_1$ and $y_2$ are the $y$-coordinates of the intersection points $M$ and $N$.

Since the focal chord $L$ of the parabola $y^2=4x$ passes through its focus $(1,0)$, its equation is $y=m(x-1)$. This line is tangent to the hyperbola $\frac{x^2}{4}-\frac{y^2}{4}=1$, and applying the tangency condition $c^2=a^2m^2-b^2$ with $c=-m$, $a^2=4$, and $b^2=4$ leads to $m^2=4m^2-4$, resulting in $m=\frac{2}{\sqrt{3}}$. Substituting this slope into the line equation and finding the intersection with the parabola yields the quadratic $y^2-2\sqrt{3}y-4=0$ for the $y$-coordinates of $M$ and $N$.

The roots of the equation $y^2-2\sqrt{3}y-4=0$ represent the $y$-coordinates $y_1$ and $y_2$, and since the product is negative, they have opposite signs, meaning their sum of absolute values $|y_1|+|y_2|$ equals the absolute difference $|y_1-y_2|$. Using the relation $|y_1-y_2|=\sqrt{(y_1+y_2)^2-4y_1y_2}$, this difference is $\sqrt{(2\sqrt{3})^2-4(-4)}=\sqrt{12+16}=2\sqrt{7}$. Substituting this value into the area expression $\frac{1}{2} \cdot 2\sqrt{2} \cdot 2\sqrt{7}$ yields a final area of $2\sqrt{14}$.

The family of circles tangent to a specific line at a given point can be expressed by combining the squared differences of the point's coordinates with the equation of the tangent line multiplied by a parameter. For the parabola defined by $y=x^2$, the slope of the tangent at the point $(2,4)$ is found by taking the derivative $\frac{dy}{dx}=2x$, which results in a slope of $4$ at $x=2$. Using the point-slope form, the equation of the tangent line becomes $y-4=4(x-2)$, which simplifies to $4x-y-4=0$.

Because the circle touches the parabola at $(2,4)$, its equation belongs to the family represented by $(x-2)^2+(y-4)^2+\lambda(4x-y-4)=0$. To find the specific circle passing through the point $(0,6)$, we substitute these coordinates into the family equation to solve for $\lambda$. This gives $(0-2)^2+(6-4)^2+\lambda(4(0)-6-4)=0$, which simplifies to $4+4-10\lambda=0$, leading to $\lambda=\frac{4}{5}$.

Substituting $\lambda=\frac{4}{5}$ back into the equation yields $(x^2-4x+4)+(y^2-8y+16)+\frac{4}{5}(4x-y-4)=0$. When expanding and collecting terms, the equation takes the form $x^2+y^2+(-4+\frac{16}{5})x + (-8-\frac{4}{5})y + (20-\frac{16}{5})=0$. By comparing this to the general circle equation $x^2+y^2+Ax+By+C=0$, we identify $A=-4+\frac{16}{5}$ and $C=20-\frac{16}{5}$. Summing these coefficients, the fractional components cancel out, leaving $A+C = -4+20=16$.

The parabola $y^2 = 6x$ follows the standard form $y^2 = 4ax$, which identifies the parameter $a = \frac{3}{2}$. A normal line to a parabola at a point $P$ with parameter $t$ is represented by the equation $y = -tx + 2at + at^3$. By substituting the coordinates $(5, -8)$ through which the normal passes, along with the calculated value of $a$, we arrive at the equation $-8 = -5t + 3t + \frac{3}{2}t^3$, which simplifies to $3t^3 - 4t + 16 = 0$. Solving this cubic equation for $t$ reveals that $t = -2$ is the valid parameter.

To find the point $Q$, we observe that the directrix of the parabola is defined by the line $x = -a = -\frac{3}{2}$. The tangent line at parameter $t$ follows the equation $yt = x + at^2$. Substituting $x = -\frac{3}{2}$ and $t = -2$ into this tangent equation gives $y(-2) = -\frac{3}{2} + \frac{3}{2}(-2)^2$, which simplifies to $-2y = -\frac{3}{2} + 6 = \frac{9}{2}$. Dividing by $-2$ results in an ordinate of $y = -\frac{9}{4}$.

The parabola $y^2 = 2x$ features a vertex at $(0,0)$ and a focus at $(1/2, 0)$. Since the circle passes through these two points, its center must lie on the vertical line $x = 1/4$, which is the perpendicular bisector of the segment connecting the vertex and focus. By defining the center as $(1/4, y_c)$ and applying the radius of 2, the circle equation becomes $(x - 1/4)^2 + (y - y_c)^2 = 4$. Substituting the point $(0,0)$ into this equation yields $(0 - 1/4)^2 + (0 - y_c)^2 = 4$, which simplifies to $1/16 + y_c^2 = 4$, confirming that $y_c^2 = 63/16$.

The parabola $y = (x - 1/4)^2 + \alpha$ possesses a vertex at $(1/4, \alpha)$. Because both the circle and the parabola share the same axis of symmetry $x = 1/4$, tangency between them occurs at this vertex. The distance between the circle's center $(1/4, y_c)$ and the parabola's vertex $(1/4, \alpha)$ must equal the radius of 2, so $|\alpha - y_c| = 2$. Squaring this relationship gives $(\alpha - y_c)^2 = 4$, and based on the geometry of the tangency, this simplifies to $(\alpha - 2)^2 = y_c^2$. Substituting $y_c^2 = 63/16$ into this result leads to $(\alpha - 2)^2 = 63/16$, and multiplying both sides by 16 produces $16(\alpha - 2)^2 = 63$, which is equivalent to $(4\alpha - 8)^2 = 63$.

Any circle tangent to a parabola at a specific point can be described by a family of circles based on the tangent line at that point. By finding the derivative of the parabola $75 x^{2}=64(5 y-3)$ and evaluating it at the point $\left(\frac{8}{5}, \frac{6}{5}\right)$, we determine that the tangent line equation is $3 x-4 y=0$. The family of circles tangent to the parabola at this point is expressed as $\left(x-\frac{8}{5}\right)^2+\left(y-\frac{6}{5}\right)^2+\lambda(3 x-4 y)=0$, where $\lambda$ represents a parameter that distinguishes each specific circle in the family.

Since the required circles must also be tangent to the $y$-axis, we set $x=0$ in the circle equation, which simplifies to the quadratic form $y^2-2 y\left(2\lambda+\frac{6}{5}\right)+4=0$. For the circle to touch the $y$-axis, this quadratic equation must have exactly one root, meaning its discriminant must be zero. Setting the discriminant to zero leads to the condition $\left(2\lambda+\frac{6}{5}\right)^2=4$, which yields two possible values for $\lambda$, specifically $\frac{2}{5}$ and $-\frac{8}{5}$.

The radius of each circle is the distance from its center to the tangent line at the $y$-axis, which corresponds to the absolute value of the x-coordinate of the center. Using the x-coordinate formula $x_c=\frac{8}{5}-\frac{3}{2}\lambda$, we find that for $\lambda=\frac{2}{5}$, the radius is $1$, and for $\lambda=-\frac{8}{5}$, the radius is $4$. Because the diameter of a circle is twice its radius, the sum of the diameters is $2(1)+2(4)$, resulting in a total value of 10.

The intersection point of tangents to a parabola is determined by the $T=0$ condition applied to the equation of the curve. By rewriting the given equation $y^{2} - 2x - 2y = 1$ in the form $(y-1)^2 = 2(x+1)$, we establish a clear reference frame to derive the equations of the tangents at the specified coordinates.

The tangent at $A(1, 3)$ is given by the equation $x = 2y - 5$, while the tangent at $B(1, -1)$ is defined by $x = -2y - 1$. Solving these two linear equations simultaneously yields the intersection point $P(-3, 1)$. Because the points $A$ and $B$ lie on the vertical line $x=1$, they form a vertical base for the triangle with a length of 4 units. Since the horizontal perpendicular distance from $P(-3, 1)$ to this base is 4 units, the total area is given by $\frac{1}{2} \times 4 \times 4 = 8$.

The parabola $2y^2 = -x$ can be rewritten in the standard form $y^2 = -\frac{1}{2}x$, which corresponds to $4a = -\frac{1}{2}$ and thus $a = -\frac{1}{8}$. A line with slope $m$ tangent to this parabola follows the equation $y = mx + \frac{a}{m}$, and since these tangents pass through the point $(2,0)$, we substitute these coordinates to obtain $0 = 2m - \frac{1}{8m}$. Solving this equation leads to $16m^2 = 1$, which determines the slopes of the tangent lines to be $m = \pm \frac{1}{4}$.

Substituting these slopes back into the tangent equation provides the two lines $y = \frac{1}{4}x - \frac{1}{2}$ and $y = -\frac{1}{4}x + \frac{1}{2}$. Rearranging these into the general linear form $x \mp 4y - 2 = 0$, we have the equations $x - 4y - 2 = 0$ and $x + 4y - 2 = 0$. Because these lines are also tangent to the circle $(x-5)^2 + y^2 = r$, the perpendicular distance from the center of the circle, $(5,0)$, to either line must be equal to the radius $\sqrt{r}$.

Applying the perpendicular distance formula, we find $\sqrt{r} = \frac{|1(5) \pm 4(0) - 2|}{\sqrt{1^2 + (\pm 4)^2}} = \frac{3}{\sqrt{17}}$. Squaring both sides of this equality yields $r = \frac{9}{17}$, and multiplying this value by 17 results in $17r = 9$.

Any line tangent to the parabola $y^2=x$ must satisfy the equation $y = mx + \frac{1}{4m}$. For this line to also be tangent to the circle $x^2+y^2=2$, the perpendicular distance from the center $(0,0)$ to the line must equal the radius $\sqrt{2}$, which is represented by the equation $\left| \frac{\frac{1}{4m}}{\sqrt{1+m^2}} \right| = \sqrt{2}$. Squaring both sides of this expression results in the biquadratic equation $32m^4 + 32m^2 - 1 = 0$, which can be treated as a quadratic in terms of $m^2$. Applying the quadratic formula leads to $m^2 = \frac{-32 \pm \sqrt{1024 + 128}}{64}$, which simplifies to the positive value $m^2 = \frac{3\sqrt{2}-4}{8}$ since $m^2$ must be real and positive. Given that $m_1$ and $m_2$ are the slopes corresponding to the two possible common tangents, their product $m_1 m_2$ is equal to $-\frac{3\sqrt{2}-4}{8}$, which results in a value of $3\sqrt{2}-4$ for the expression $8 |m_1 m_2|$.

To find the intersection of the common tangents, consider a line $y=mx+c$ that is tangent to both the parabola $y^2=4x$ and the circle $x^2+y^2=\frac{9}{4}$. The condition for tangency to the parabola is $c=\frac{1}{m}$, and the condition for tangency to the circle is $c^2=r^2(1+m^2)$, where $r=\frac{3}{2}$. Substituting the parabola's condition into the circle's condition yields $\frac{1}{m^2}=\frac{9}{4}(1+m^2)$, which simplifies to the equation $9m^4+9m^2-4=0$. Factoring this quadratic in terms of $m^2$ gives $(3m^2-1)(3m^2+4)=0$, and since $m^2$ must be positive, we have $m^2=\frac{1}{3}$. This implies the tangent lines are $y=\pm\frac{1}{\sqrt{3}}x\pm\sqrt{3}$, which intersect at the point $Q(-3, 0)$.

The ellipse is centered at the origin with a semi-minor axis equal to the distance $OQ$, which is 3, and a given semi-major axis of 6. Using the standard ellipse relation $b^2=a^2(1-e^2)$ with $b=3$ and $a=6$, we find $9=36(1-e^2)$, which simplifies to $1-e^2=\frac{1}{4}$, or $e^2=\frac{3}{4}$. The length of the latus rectum $l$ is calculated as $l=\frac{2b^2}{a}$, substituting the values gives $l=\frac{2(9)}{6}=3$. The final value is determined by the ratio $\frac{l}{e^2}$, which is 3 divided by $\frac{3}{4}$, resulting in 4.

The parabola $y^2=4x$ has its focus at $(1,0)$, and any focal chord endpoints can be parameterized by $t$ as $(t^2, 2t)$ and $(\frac{1}{t^2}, -\frac{2}{t})$. For the segment $PQ$ to subtend an angle of $\frac{\pi}{2}$ at the fixed point $(3,0)$, the product of the slopes of the segments connecting this point to $P$ and $Q$ must equal $-1$. This calculation leads to the condition $t^2=1$, which identifies the chord as the vertical line $x=1$ with endpoints $(1,2)$ and $(1,-2)$.

Since this line segment $PQ$ is also a focal chord of the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
the ellipse's focus $(ae,0)$ must lie on the line $x=1$, yielding $ae=1$ or $a^2=\frac{1}{e^2}$. Substituting $x=1$ into the ellipse equation results in $\frac{1}{a^2}+\frac{4}{b^2}=1$, which transforms into the following expression by replacing $b^2$ with $a^2(1-e^2)$:
$1+\frac{4}{1-e^2}=\frac{1}{e^2}$
Rearranging this equality produces the biquadratic equation
$e^4-6e^2+1=0$
and solving for the reciprocal $\frac{1}{e^2}$ provides the final result $3+2\sqrt{2}$.

Because the line $y=4+kx$ is tangent to the parabola $y=x-x^2$, the intersection points occur where $4+kx = x-x^2$, resulting in the quadratic equation $x^2+(k-1)x+4=0$. For the line to be a tangent, this equation must have exactly one root, meaning its discriminant must be zero; thus $(k-1)^2 - 16 = 0$, which leads to $k-1 = \pm 4$. Given the condition $k>0$, we determine that $k=5$.

Substituting $k=5$ into the quadratic equation yields $x^2+4x+4=0$, which simplifies to $(x+2)^2=0$, identifying the $x$-coordinate of the point of tangency $P$ as $-2$ and its $y$-coordinate as $-6$. The vertex $V$ of the parabola $y=-x^2+x$ is located at $x = -b/(2a) = -1/(2(-1)) = 1/2$, and substituting this into the parabola's equation gives a $y$-coordinate of $1/4$. The slope of the line passing through $P(-2, -6)$ and $V(1/2, 1/4)$ is found by calculating $\frac{1/4 - (-6)}{1/2 - (-2)}$, which simplifies to $\frac{25/4}{5/2}$, resulting in a final value of $\frac{5}{2}$.

To find the slopes of the tangents, we use the angle between lines formula $\tan \theta = \left|\frac{m - 3}{1 + 3m}\right|$ with $\theta = \frac{\pi}{4}$, which simplifies to $\frac{m - 3}{1 + 3m} = \pm 1$ and yields $m = -2$ and $m = \frac{1}{2}$. Using the general contact point formula $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$ for the parabola $y^2 = 4ax$, we determine the points $A\left(\frac{a}{4}, -a\right)$ and $B(4a, 4a)$, while the focus is located at $S(a, 0)$. Comparing the slopes of $AS$ and $SB$, we calculate $\frac{-a - 0}{\frac{a}{4} - a} = \frac{4}{3}$ and $\frac{4a - 0}{4a - a} = \frac{4}{3}$ respectively. Since these slopes are identical, the points $A, B$, and $S$ are collinear for any $a > 0$.

A common tangent to two curves exists where a line shares the same slope $m$ and y-intercept for both functions. The tangent to the parabola $y=x^2$ is defined by $y=mx-\frac{1}{4}m^2$, while the tangent to the parabola $y=-(x-2)^2$ is given by $y=m(x-2)+\frac{1}{4}m^2$. Setting these two expressions equal to each other yields $mx-\frac{1}{4}m^2 = mx-2m+\frac{1}{4}m^2$, which simplifies to $2m = \frac{1}{2}m^2$. Solving this equation results in slopes of $m=0$ and $m=4$, and substituting $m=4$ into the tangent expression produces $y=4x-4$, which simplifies to $y=4(x-1)$.

The minimum distance between a curve and a circle is realized along the normal line passing through the center of the circle. At any point $(x_0, x_0^2)$ on the parabola $y=x^2$, the normal has a slope of $-\frac{1}{2x_0}$, leading to the equation $y - x_0^2 = -\frac{1}{2x_0}(x - x_0)$. Substituting the center $(1, -1)$ into this equation yields $-1 - x_0^2 = -\frac{1}{2x_0}(1 - x_0)$, which simplifies to the cubic equation $2x_0^3 + 3x_0 - 1 = 0$. Testing the function $f(x) = 2x^3 + 3x - 1$ at the boundaries gives $f\left(\frac{1}{4}\right) = -\frac{7}{32}$ and $f\left(\frac{1}{2}\right) = \frac{3}{4}$, confirming that the abscissa for the point $Q$ on the parabola lies within the interval $\left(\frac{1}{4}, \frac{1}{2}\right)$, where the point $P$ on the circle also resides.