# Tangent to a Parabola

## Equation of the Tangent to the Parabola $y^2=4ax$ at a Point $(x_1,y_1)$ on the Parabola

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ be any two points on a parabola $y^2=4ax$ such that the point $Q$ is near to the point $P$.

Then, $y_1^2=4ax_1$ $y_2^2=4ax_2$

Subtracting these equations, $y_2^2-y_1^2=4a(x_2-x_1)$ $\frac{y_2-y_1}{x_2-x_1}=\frac{4a}{y_2+y_1}\text{ __(i)}$

Also, the equation of line $PQ$ is, $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

From $\text{(i)}$ and $\text{(ii)}$, $y-y_1=\frac{4a}{y_2+y_1}(x-x_1)$

Now, the line $PQ$ becomes the tangent at $P$ when $Q$ approaches $P$, i.e. when $x_2\to x_1$ and $y_2\to y_1$. Hence, the equation of the tangent at $P$ is, $y-y_1=\frac{4a}{2y_1}(x-x_1)$ $yy_1-y_1^2=2ax-2ax_1$ $yy_1-4ax_1=2ax-2ax_1$ $\therefore yy_1=2a(x+x_1)$

Similarly, the equation of the tangent to the parabola $x^2=4ay$ at the point $(x_1,y_1)$ is $xx_1=2a(y+y_1)$.

## Condition of Tangency of a Straight Line to a Parabola and Equation of Tangent in Slope Form

Let the equation of a parabola be $y^2=4ax$ and, the equation of a line be $y=mx+c$ Now, let us solve these two equations simultaneously to obtain the points of intersection of the parabola and the line. For this, eliminating $y$, $(mx+c)^2=4ax$ $m^2x^2+2mcx+c^2-4ax=0$ $m^2x^2+2(mc-2a)x+c^2=0\text{ __(1)}$

This is quadratic in $x$. Hence, $x$ has two values and corresponding to the two values of $x$, we can obtain the two values of $y$. In this way, we can obtain the points of intersection.

Now, the line will be a tangent to the parabola if it intersects the parabola in two coincident points i.e. if the roots are real and equal. $\therefore B^2-4AC=0$ $4(mc-2a)^2-4m^2c^2=0$ $16a(a-mc)=0$ $\therefore c=\frac{a}{m}$ This is the condition of tangency. Thus, the equation of a line tangent to the parabola is given by, $y=mx+\frac{a}{m}\text{ __(1)}$ This equation of the tangent is in slope form.

Again, the equation of the tangent to the parabola $y^2=4ax$ at the point $(x_1,y_1)$ is given by $yy_1=2a(x+x_1)\text{ __(2)}$

Hence, the equations $\text{(1)}$ and $\text{(2)}$ will represent a same line if $\frac{1}{y_1}=\frac{m}{2a}=\frac{a/m}{2ax_1}$ By solving this, we get, $x_1=\frac{a}{m^2}\:\text{ and }\:y_1=\frac{2a}{m}$ Thus, the point of contact is $\left(\frac{a}{m^2},\frac{2a}{m}\right)$.

## Equation of the Tangent to the Parabola in Parametric Form

Let $P(at_1^2,2at_1)$ and $Q(at_2^2,2at_2)$ be two points on the parabola $y^2=4ax$. Then the equation of the line joining the two points $P$ and $Q$ is $y-2at_1=\frac{2at_2-2at_1}{at_2^2-at_1^2}(x-at_1^2)$ $y-2at_1=\frac{2}{t_2+t_1}(x-at_1^2)$

Simplifying this equation, we get $y(t_1+t_2)=2x+2at_1t_2\text{ __(1)}$ When $Q\to P$ $(t_2\to t_1)$, then the equation $\text{(1)}$ becomes $yt_1=x+a_1t_1^2$ $x-yt_1+at_1^2=0\text{ __(2)}$

This is the equation of the tangent in parametric form.

Cor. Equation $\text{(1)}$ gives the equation of the chord joining the points $P$ and $Q$. If this chord passes through the focus $(a,0)$ then $0=2a+2at_1t_2$ $\therefore t_1t_2=-1$

### Obtain the equation of a tangent to a parabola $y^2=8x$ at $(2,-4)$.

Equation of the parabola is $y^2=8x$ $\therefore 4a=8\Rightarrow a=2$

Given point is $(x_1,y_1)$$=(2,-4)$. Then the equation of the tangent is $yy_1=2a(x+x_1)$ $-4y=4(x+2)$ $4x+4y+8=0$ $\therefore x+y+2=0$

### Obtain the equation of a tangent to a parabola $y^2=3x$ parallel to the line $y=2x+1$.

Equation of the parabola is $y^2=3x$ $\therefore 4a=3\Rightarrow a=\frac{3}{4}$ Given line is $y=2x+1$ $\text{Its Slope}=2$

Since the required tangent line is parallel to the given line, $\text{Slope of tangent }(m)=2$

Hence, the equation of the tangent is $y=mx+\frac{a}{m}$ $y=2x+\frac{3/4}{2}$ $\therefore 16x-8y+3=0$

### Prove that the line $lx+my+n=0$ touches the parabola $y^2=4ax$ if $ln=am^2$.

Given line is $lx+my+n=0$ and, the equation of parabola is $y^2=4ax$ Eliminating$x$ from above two equations, $l\frac{y^2}{4a}+my+n=0$ $ly^2+4amy+4an=0$

This equation is quadratic in $y$. Now, the line $lx+my+n=0$ will touch the parabola if discriminant of this quadratic equation is zero. $\text{i.e. }B^2-4AC=0$ $16a^2m^2-16aln=0$ $am^2-ln=0$ $\therefore ln=am^2$