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# Tangent to a Parabola - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Tangents of Parabola in Point Form, Tangents of Parabola in Slope Form is considered one of the most asked concept.

• 100 Questions around this concept.

## Solve by difficulty

If the line $\mathrm{x+y-1=0}$ touches the parabola $\mathrm{y^{2}=k x}$, then the value of $\mathrm{k}$ is.

The slope of the line touching both the parabolas y2= 4x and x2 = -32y is:

Tangents are drawn to the parabola $\mathrm{Y}^{2}=4 \mathrm{ax}$ at the point $\mathrm{A}$ and $\mathrm{B}$  intersect at $\mathrm{C}$. If '$\mathrm{S}$' be the focus of the parabola then, SA, SC and SB forms

## Concepts Covered - 0

Tangents of Parabola in Point Form

Tangents of Parabola in  Point Form

$\\ {\text {Equation of the tangent to the parabola } \mathrm{y}^{2}=4 \mathrm{ax} \text { at the point } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { is }} \\ {\mathrm{y} \mathrm{y}_{1}=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_{1}\right)}$

$\\\text{The given equation is }\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;y^2=4ax}\\\text{Differentiating with respect to x, we get}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;2y\frac{dy}{dx}=4a}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\frac{dy}{dx}=\frac{2a}{y}}\\\\\text { Now } m=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{2 a}{y_{1}}\\\\\text{Equation of tangent at point }(x_1,y_1)\\\\\mathrm{\Rightarrow \;\;\;\;\;\left ( y-y_1 \right )=\frac{2a}{y_1}\left ( x-x_1 \right )}\\\\\mathrm{\Rightarrow \;\;\;\;\;y y_{1}-y_{1}^{2}=2 a x-2 a x_{1}} \\\\\mathrm{\Rightarrow \;\;\;\;\;y y_{1}=2 a x-2 a x_{1}+4 a x_{1}} \\\\\mathrm{\Rightarrow \;\;\;\;\,\,y y_{1}=2 a \left ( x+x_1 \right )}$

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

Note:

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

For example, the tangent to 4y = x2 + 2x - 9 at (x1,y1) is 2(y+y1) = xx1 + (x+x1) - 9

Tangents of Parabola in Parametric Form

Tangents of Parabola in Parametric Form

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

Proof:
$\\ {\text {Equation of the tangent to the parabola } \mathrm{y}^{2}=4 \mathrm{ax} \text { at the point } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { is }} \\ {\mathrm{yy}_{1}=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_{1}\right)} \\ {\text { replace } \mathrm{x}_{1} \rightarrow \mathrm{at}^{2}, \mathrm{y}_{1} \rightarrow 2 \mathrm{at}} \\ {\mathrm{y}(2 \mathrm{at})=2 \mathrm{a}\left(\mathrm{x}+\mathrm{at}^{2}\right) \Rightarrow \mathrm{yt}=\mathrm{x}+\mathrm{at}^{2}}$

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

Tangents of Parabola in Slope Form

Tangents of Parabola in Slope Form

$\\ {\text {Equation of the tangent to the parabola } \mathrm{y}^{2}=4 \mathrm{ax} \text { at the point } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { is }} \\ {\mathrm{y} \mathrm{y}_{1}=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_{1}\right)}$

If m is the slope of the tangent, then

$\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{m}=\frac{2 \mathrm{a}}{\mathrm{y}_{1}} \Rightarrow \mathrm{y}_{1}=\frac{2 \mathrm{a}}{\mathrm{m}}} \\ {\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { lies on the parabola } \mathrm{y}^{2}=4 \mathrm{ax}} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{y}_{1}^{2}=4 \mathrm{ax}_{1} \Rightarrow \frac{4 \mathrm{a}^{2}}{\mathrm{m}^{2}}=4 \mathrm{ax}_{1}} \\ {\therefore\;\;\;\;\;\;\;\;\;\;\; \mathrm{x}_{1}=\frac{\mathrm{a}}{\mathrm{m}^{2}}} \\ {\text { put the value of } \mathrm{x}_{1} \text { and } \mathrm{y}_{1} \text { in the equation } \mathrm{yy}_{1}=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_{1}\right)} \\ {\text { we get }} \\ {\Rightarrow \;\;\;\;\;\;\;\;\;\;\mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}} \\ {\text { which is equation of tangent of the parabola in slope form. }}$
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

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

Note:

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

Point of Intersection of Tangent

Point of Intersection of Tangent

$\\ {\text {Two points, } P \equiv\left(a t_{1}^{2}, 2 a t_{1}\right) \text {and } Q \equiv\left(a t_{2}^{2}, 2 a t\right) \text {on the parabola } y^{2}=4 a x .} \\ {\text {Then, equation of tangents at } P \text { and } Q \text { are }} \\ {t_{1} y=x+a t_{1}^{2}} \\ {t_{2} y=x+a t_{1}^{2}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots() \\ {\text {Solving (i) and (ii)}} \\ {\text {we get, } x=a t_{1} t_{2}, y=a\left(t_{1}+t_{2}\right)} \\ {\text {Point of Intersection of tangents drawn at point } P \text { and } Q \text { is }} \\ {\left(a t_{1} t_{2}, a\left(t_{1}+t_{2}\right)\right)}$

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

$\mathbf{\left(a t_{1} t_{2}, \;a\left(t_{1}+t_{2}\right)\right)}$

TIP

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

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