Tangent to a Parabola - Practice Questions & MCQ

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 $\mathrm{yy}_1=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)$

The given equation is

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

$
\begin{aligned}
& 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{aligned}
$
Equation of tangent at point $\left(x_1, y_1\right)$

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

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 $2\left(y+y_1\right)=x x_1+\left(x+x_1\right)-9$

Tangents of Parabola in Parametric Form

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

Proof:
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
$

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)
$
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 $\mathrm{x}_1$ and $\mathrm{y}_1$ in the equation $\mathrm{yy}_1=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)$ 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 point of contact are $\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$

Slope form of tangent for other forms of parabola

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, equation of tangents at $P$ and $Q$ are

$
\begin{aligned}
& t_1 y=x+a t_1^2 \\
& t_2 y=x+a t_1^2
\end{aligned}
$
Solving (i) and (ii)

$
\text { 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)
$

Point of Intersection of tangents drawn at point P and Q is

$
\left(\mathbf{a t}_1 \mathbf{t}_{\mathbf{2}}, \mathbf{a}\left(\mathbf{t}_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.