Tangent to a Parabola

A line that touches the parabola exactly at one point is called the tangent to a parabola.

Condition for Tangency
1. The line $y=m x+c$ is a tangent to the parabola $y^2=4 a x$, if $c=a / m$.
2. The line $y=m x+c$ is a tangent to the parabola $x^2=4 a y$, if $c=-a m^2$

Equation of Tangents of Parabola in Point Form

Equation of the tangent to the parabola $y^2=4 a x$ at the point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{yy}_1=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)$

Derivation of Equation of Tangents of Parabola in Point Form

The given equation is

$
y^2=4 a x
$
Differentiating with respect to $x$, we get

$
\begin{gathered}
2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=4 \mathrm{a} \\
\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{\mathrm{y}} \\
\text { Now } m=\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{2 a}{y_1}
\end{gathered}
$
Equation of tangent at point $\left(x_1, y_1\right)$

$
\begin{array}{ll}
\Rightarrow & \left(\mathrm{y}-\mathrm{y}_1\right)=\frac{2 \mathrm{a}}{\mathrm{y}_1}\left(\mathrm{x}-\mathrm{x}_1\right) \\
\Rightarrow & \mathrm{yy}_1-\mathrm{y}_1^2=2 \mathrm{ax}-2 \mathrm{ax}_1 \\
\Rightarrow & \mathrm{yy}_1=2 \mathrm{ax}-2 \mathrm{ax}_1+4 \mathrm{ax}_1 \\
\Rightarrow \quad & \mathrm{yy}_1=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)
\end{array}
$

The equation of tangents to all standard parabolas in terms of the parameter of the point of contact and the slope of the tangents are tabulated as shown below:

$\begin{array}{c||c c} \\ \mathbf { Equation \;of \;Parabola } & {\mathbf { A \;tangent\; at\; } P\left(x_{1}, y_{1}\right)} \\ \\ \hline \hline\\y^{2}=4ax & {y y_{1}=2 a\left(x+x_{1}\right)} & {} \\\\ {y^{2}=-4 a x} & {y y_{1}=-2 a\left(x+x_{1}\right)} & {} \\\\ {x^{2}=4 a y} & {x x_{1}=2 a\left(y+y_{1}\right)} & {} \\\\ {x^{2}=-4 a y} & {x x_{1}=-2 a\left(y+y_{1}\right)} & {} \\ \end{array}$

Note:

The same procedure can be applied to any general equation of parabola as well

For example, the tangent to $4 y=x^2+2 x-9$ at $\left(x_1, y_1\right)$ is $2\left(y+y_1\right)=x x_1+\left(x+x_1\right)-9$

Equation of Tangents of Parabola in Parametric Form

The equation of tangent to the parabola $y^2=4 \mathrm{ax}$ at the point $\left(\mathrm{at}^2, 2 \mathrm{at}\right)$ is ty $=x+a t^2$

Derivation of Equation of Tangents of Parabola in Parametric Form

Equation of the tangent to the parabola $y^2=4 a x$ at the point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{yy}_1=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)$ replace $\mathrm{x}_1 \rightarrow \mathrm{at}^2, \mathrm{y}_1 \rightarrow 2$ at

$
y(2 a t)=2 a\left(x+a t^2\right) \Rightarrow y t=x+a t^2
$

$\begin{array}{c||c cl} \\\mathbf { {Equation \;of \;Parabola} } & {\mathbf { Coordinate }} & {\mathbf { Tangent\; Equation }}\\ \\ \hline\hline\\ {\color{Teal} y^{2}=4ax} & {\color{Teal} {\left(at^{2}, 2 a t\right)}} & {\color{Teal} {t y=x+a t^{2}}} \\ \\ {\color{Red} x^{2} {=4 a y}} & {\color{Red} {(2 a t, a t^2)}} & {\color{Red} {t x=y+a t^{2}}} \\ \\ {\color{Teal} y^{2}{=-4 a x}} & {\color{Teal} {\left(-a t^{2}, 2 a t\right)} }& {\color{Teal} {t y=-x+a t^{2}}} \\ \\ {\color{Red} x^{2} {=} {-4 a y}} & {\color{Red} {\left(2 a t,-at^{2}\right)}} & {\color{Red} {t x=-y+a t^{2}} }\\ \end{array}$

Equation of Tangents of Parabola in Slope Form

Equation of the tangent to the parabola $y^2=4 a x$ at the point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{yy}_1=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)$

Derivation of Equation of Tangents of Parabola in Slope Form

If $m$ is the slope of the tangent, then

$
\mathrm{m}=\frac{2 \mathrm{a}}{\mathrm{y}_1} \Rightarrow \mathrm{y}_1=\frac{2 \mathrm{a}}{\mathrm{m}}
$

$\left(\mathrm{x}_1, \mathrm{y}_1\right)$ lies on the parabola $\mathrm{y}^2=4 \mathrm{ax}$

$
\begin{aligned}
\mathrm{y}_1^2 & =4 \mathrm{ax}_1 \Rightarrow \frac{4 \mathrm{a}^2}{\mathrm{~m}^2}=4 \mathrm{ax}_1 \\
\therefore \quad \mathrm{x}_1 & =\frac{\mathrm{a}}{\mathrm{m}^2}
\end{aligned}
$

put the value of $x_1$ and $y_1$ in the equation $y_1=2 a\left(x+x_1\right)$ we get

$
\Rightarrow \quad y=m x+\frac{a}{m}
$

which is the equation of the tangent of the parabola in slope form.
The coordinates of the point of contact are $\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$

Slope form of tangent for other forms of parabola

$\begin{array}{c||cc} \mathbf { Equation \;of \;Parabola } & {\mathbf { Point \;of \;Contact }} & {\mathbf { Tangent\; Equation }} \\\\ \hline\hline \\ {\color{Black} y^{2}{=4 a x}} & {\color{Black} {\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)}} & {\color{Black} {y=m x+\frac{a}{m}}}\\ \\ {\color{Black} y^{2}{=-4 a x}} & {\color{Black} {\left(-\frac{a}{m^{2}}, \frac{2 a}{m}\right)}} & {\color{Black} {y=m x-\frac{a}{m}}} \\\\ {\color{Black} x^{2}{=4 a y}} & {\color{Black} {\left(2am, am^2\right)}} & {\color{Black} {y=m x-am^2}} \\\\ {\color{Black} x^{2}{=-4 a y}} & {\color{Black} {\left(2am, -am^2\right)}} & {\color{Black} {y=m x+am^2}} \end{array}$

Note:

When the vertex of the shifted parabola is at (h, k), then replace x by (x-h) and y by (y-k) in the equations of tangents.

Point of Intersection of Tangent

Two points, $P \equiv\left(a t_1^2, 2 a t_1\right)$ and $Q \equiv\left(a t_2^2, 2 a t\right)$ on the parabola $y^2=4 a x$.
Then, the equation of tangents at $P$ and $Q$ are
$t_1 y=x+a t_1^2$
$t_2 y=x+a t_1^2$
Solving (i) and (ii)
we get, $x=a t_1 t_2, y=a\left(t_1+t_2\right)$
Point of Intersection of tangents drawn at point $P$ and $Q$ is $\left(a t_1 t_2, a\left(t_1+t_2\right)\right)$

The point of Intersection of tangents drawn at points $P$ and $Q$ is $\left(\mathbf{a} \mathbf{t}_1 \mathbf{t}_{\mathbf{2}}, \mathbf{a}\left(\mathbf{t}_{\mathbf{1}}+\mathbf{t}_{\mathbf{2}}\right)\right)$

TIP:

The locus of the point of intersection of the mutually perpendicular tangents to a parabola is the directrix of the parabola.

All the above forms of the equation of tangent can be obtained by the calculus method also in which we differentiate the equation of parabola to get the slope of the tangent to parabola at any point on it.

Equation of tangent to any parabola

To find the equation of the tangent to a standard parabola, we use the standard equation of tangent. But if the equation of a parabola is not in standard form, we use the calculus method to find the equation of tangent. In this method, we differentiate the equation of the parabola to find the slope of tangent or point of contact on the parabola.

Properties of tangent

1) In any parabola, the foot perpendicular from focus upon any tangent lies on the tangent at the vertex.

2) In any parabola, the image of focus in any tangent lies on the directrix.

3) In any parabola, tangent at any point P on it bisects the angle between the focal chord through P and the perpendicular from P to the directrix.

4) In any parabola, the length of tangent between the point of contact on the curve and the point where it meets the directrix subtends the right angle at focus.

5) Tangents at extremities of a focal chord are perpendicular and intersect on the directrix. In other words, the locus of point of intersection of tangents at the extremities of the focal chord of the parabola is directrix. Or Chord of contact w.r.t. any point on the directrix of the parabola is a focal chord.

Solved Examples Based on Tangents of Parabola

Example 1: A triangle is formed by the tangents at the point $(2,2)$ on the curves $y^2=2 x$ and $x^2+y^2=4 x$, and the line $x+y+2=0$. If $r$ is the radius of its circumcircle, then $r^2$ is equal to
[JEE MAINS 2023]
Solution: Tangent at $y^2=2 x$

$
\begin{aligned}
& T: 2 y=2\left(\frac{x+2}{2}\right) \\
& 2 y=x+2
\end{aligned}
$
Tangent at $x^2+y^2=4 x$

$
\begin{aligned}
& 2 x+2 y=\frac{4 \times(x+2)}{2} \\
& 2 x+2 y=2 x+4 \\
& y=2
\end{aligned}
$

$
M_{P R}=-1
$
Slope of $\perp^{\mathrm{rt}}$ Bisector $=1$

$
\begin{aligned}
& \mathrm{y}-1=1(x+3) \\
& \mathrm{y}=\mathrm{x}+3+1 \\
& \mathrm{y}=\mathrm{x}+4
\end{aligned}
$

$\perp^{\mathrm{r}}$ Bisector of PQ

$
x=-1
$

$\therefore$ Centre is

$
\begin{aligned}
& y=-1+4=3 \\
& (-1,3)
\end{aligned}
$
Radius: $\mathrm{r}=\sqrt{(-1+4)^2+(3-2)^2}$

$
\begin{aligned}
& =\sqrt{9+1} \\
& =\sqrt{10} \\
& \mathrm{r}^2=10
\end{aligned}
$
Hence, the answer is 10.

Example 2: If the tangent at a point $P$ on the parabola $y^2=3 x$ is parallel to the line $x+2 y=1$ and the tangents at the points $Q$ and $R$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ are perpendicular to the line $x-y=2$, then the area of the triangle PQR is:
[JEE
MAINS 2023]
Solution

$\begin{array}{ll}x+2 y=1 & y^2=3 x \\ \mathrm{~m}=-\frac{1}{2} & \mathrm{~T}_{\mathrm{p}}: y=-\frac{1}{2} \mathrm{x}+\frac{\frac{3}{4}}{-\frac{1}{2}}\end{array}$
$
\begin{aligned}
& y=-\frac{x}{2}-\frac{3}{2} \\
& 2 y+x+3=0
\end{aligned}
$

$\begin{array}{ll}x-y=2 & E: \frac{x^2}{4}+\frac{y^2}{1}=1 \\ m=1 & y=-x \pm \sqrt{(-1)^2 4+1}\end{array}$

the slope of the tangent at $Q \& R$ is -1

$
\begin{aligned}
& y=-x \pm \sqrt{5} \\
& x+y=\sqrt{5}
\end{aligned}
$

(2) $x+y=-\sqrt{5}$

$\begin{array}{lcr}\text { Point P: } & \text { Point Q: } & \text { Point R: } \\ \mathrm{T}=\mathrm{O} & \frac{\mathrm{xx}_2}{4}+\frac{\mathrm{yy}_2}{1}=1 & \frac{\mathrm{x}_2}{1}=\frac{4 \mathrm{y}_2}{1}=\frac{-4}{\sqrt{5}} \\ \mathrm{yy}_1=\frac{3}{2}\left(x+x_1\right) & x x_2+4 \mathrm{yy}_2-4=0 & x_2=\frac{-4}{\sqrt{5}}, \mathrm{y}=\frac{1}{\sqrt{5}} \\ 3 x-2 y_1+3 x_1=0 & \frac{x_2}{1}=\frac{4 y_2}{1}=\frac{-4}{-\sqrt{5}} & \end{array}$

$x_2=\frac{4}{\sqrt{5}}-y_2=\frac{1}{\sqrt{5}}$

Comparison with (1)

$
\begin{aligned}
& \frac{3}{1}=\frac{-2 \mathrm{y}_1}{2}=\frac{3 \mathrm{x}_1}{3} \\
& \mathrm{y}_1=-3, \quad \mathrm{x}_1=3
\end{aligned}
$
Area of $\triangle P Q R$

$
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
3 & -3 & 1 \\
\frac{4}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 1 \\
-\frac{4}{\sqrt{5}} & -\frac{1}{\sqrt{5}} & 1
\end{array}\right| \\
& \Rightarrow \frac{1}{2}\left[3\left(\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}\right)+3\left(\frac{4}{\sqrt{5}}+\frac{4}{\sqrt{5}}\right)+1\left(-\frac{4}{5}+\frac{4}{5}\right)\right] \\
& \Rightarrow \frac{1}{2}\left[\frac{6}{\sqrt{5}}+\frac{24}{\sqrt{5}}\right] \\
& \Rightarrow \frac{1}{2} \times \frac{30}{\sqrt{5}}=\frac{5 \times 3}{\sqrt{5}}=3 \sqrt{5}
\end{aligned}
$

Hence, the answer is $3 \sqrt{5}$

Example 3: Let $x^2+y^2+A x+B y+C=0$ be a circle passing through $(0,6)$ and touching the parabola $\mathrm{y}=\mathrm{x}^2$ at $(2,4)$. Then $\mathrm{A}+\mathrm{C}$ is equal to $\qquad$
[JEE MAINS 2022]

Solution

$
\begin{aligned}
& 0+36+6 B+C=0 \\
& 6 B+C=-36 \quad \cdots(i)
\end{aligned}
$
$
\begin{aligned}
& 4+16+2 A+4 B+C=0 \\
& 2 A+4 B+C=-20 \quad \cdots(\text { ii })
\end{aligned}
$

Also,

$
\begin{aligned}
& 2 \mathrm{x}+2 \mathrm{y} \mathrm{y}^{\prime}+\mathrm{A}+\mathrm{By}^{\prime}=0 \\
& \left.\mathrm{y}^{\prime}\right)_{(2,4)}=\frac{-\mathrm{A}-4}{8+\mathrm{B}}=(2 \mathrm{x})_{\mathrm{x}=2} \\
& \frac{-(\mathrm{A}+4)}{\mathrm{B}+8}=4 \\
& \mathrm{~A}+4 \mathrm{~B}=-36 \quad \cdots \text { (iii) }
\end{aligned}
$
$\begin{aligned} & \mathrm{A}=\frac{-4}{5}, \mathrm{~B}=\frac{-44}{5}, \mathrm{C}=\frac{84}{5} \\ & \therefore \mathrm{A}+\mathrm{C}=16\end{aligned}$

Hence, the answer is 16

Example 4: Let $\mathrm{P}: \mathrm{y}^2=4 \mathrm{ax}, \mathrm{a}>0$ be a parabola with focus S . Let the tangents to the $\underline{\pi}$ parabola P make an angle of $\overline{4}$ with the line $y=3 x+5$ touch the parabola $P$ at $A$ and $B$. Then the value of a for which $\mathrm{A}, \mathrm{B}$ and S are collinear is
[JEE MAINS 2022]

Solution

Tangents $\mathrm{T}_1$ and $\mathrm{T}_2$ make an angle $\frac{\pi}{4}$ with a line $\mathrm{y}=3 \mathrm{x}+5$ means that tangents are perpendicular

Using the property of a parabola, the point of contact of perpendicular tangents always forms a focal chord. So A, S, B will always be collinear for any value of a>0.

Hence the answer is any a>0

Example 5:
If a line along a chord of the circle $4 x^2+4 y^2+120 x+675=0$, passes through the point $(-30,0)$ and is tangent to the parabola $y^2=30 x$, then the length of this chord is $[J E E$ MAINS 2021]

Solution: The line is passing through ( $-30,0$ ) and is tangent to

$
y^2=4 \cdot\left(\frac{30}{4}\right) x
$
Any tangent to this parabola is

$
y=m x+\frac{30}{4 m}----(i)
$

It passes through (-30,0)
$
\begin{aligned}
& \therefore 0=-30 m+\frac{30}{4 m} \\
& \Rightarrow 30 m=\frac{30}{4 m} \\
& \Rightarrow m=\frac{1}{2}, \frac{-1}{2}
\end{aligned}
$

For $m=\frac{1}{2}: y=\frac{x}{2}+15 \Rightarrow x-2 y+30=0$
Given circle is $x^2+y^2+30 x+\frac{675}{4}=0$

Centre is $(-15,0)$
Distance of centre from line

$
\begin{aligned}
& =\frac{|-15+30|}{\sqrt{1^2+2^2}}=\frac{15}{\sqrt{5}} \\
& \text { Radius }=\sqrt{225-\frac{675}{4}}=\frac{15}{2}
\end{aligned}
$

$\begin{aligned} \therefore \text { Length of chord } & =2 \sqrt{\frac{225}{4}-\frac{225}{5}} \\ & =\frac{2 \cdot 15}{2 \sqrt{5}}=\frac{15}{\sqrt{5}} \\ & =3 \sqrt{5}\end{aligned}$

Hence, the answer is $3\sqrt{5}$

Summary

Tangents to any parabola give us an idea about the behaviour of parabola and their interaction with straight lines. Understanding tangents enhances our ability to analyze and predict the behaviour of parabolic curves, making them a fundamental concept in both theoretical mathematics and practical applications.