Tangent to a Parabola - Practice Questions & MCQ

Tangents of Parabola in  Point Form

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

The given equation is

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

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

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

Note:

The same procedure can be applied to any general equation of parabola as well
For example, the tangent to $4y=x^2+2x-9$ at $(x_1,y_1)$ is $2(y+y_1)=x{x}_1+(x+x_1)-9$

Tangents of Parabola in Parametric Form

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

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

$y(2at)=2a(x+a{t}^2)\Rightarrow yt=x+a{t}^2$

Tangents of Parabola in Slope Form

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

${\mathrm{yy}}_1=2\mathrm{a}(\mathrm{x}+{\mathrm{x}}_1)$
If $m$ is the slope of the tangent, then

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

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

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

put the value of ${\mathrm{x}}_1$ and ${\mathrm{y}}_1$ in the equation ${\mathrm{yy}}_1=2\mathrm{a}(\mathrm{x}+{\mathrm{x}}_1)$ we get

$\Rightarrow y=mx+\frac{a}{m}$

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

Slope form of tangent for other forms of parabola

Point of Intersection of Tangent

Two points, $P\equiv (a{t}_1^2,2a{t}_1)$ and $Q\equiv (a{t}_2^2,2at)$ on the parabola $y^2=4ax$.
Then, equation of tangents at $P$ and $Q$ are

$\begin{array}{rl}& t_{1}y=x+a{t}_1^2\\ & t_{2}y=x+a{t}_1^2\end{array}$
Solving (i) and (ii)

$\text{ we get, }x=a{t}_1{t}_2,y=a(t_1+t_2)$
Point of Intersection of tangents drawn at point $P$ and $Q$ is

$(a{t}_1{t}_2,a(t_1+t_2))$

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

$({\mathbf{a}\mathbf{t}}_1{\mathbf{t}}_{\mathbf{2}},\mathbf{a}({\mathbf{t}}_1+{\mathbf{t}}_{\mathbf{2}}))$
TIP
The locus of the point of intersection of the mutually perpendicular tangents to a parabola is the directrix of the parabola.