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 $P$.

Tangent to a Parabola

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. \[\text{i.e. }B^2-4AC=0\] \[16a^2m^2-16aln=0\] \[am^2-ln=0\] \[\therefore ln=am^2\]


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