Question

If the line \[lx+my+n=0\] touches the parabola \[{{y}^{2}}=4ax\] then prove that \[nl=a{{m}^{2}}\]

Answer 1

Hint: The rough figure that represents the given information is shown below.

We solve this problem by substituting the value of \['x'\] in terms of \['y'\] from the line equation in the parabola equation. We know that if the line touches a parabola that means the line and parabola have only one point of intersection.
So, by substituting the line equation in parabola we get the quadratic equation of \['y'\] where we make the discriminant equal to 0 as there is only one point of intersection.
The discriminant of \[a{{x}^{2}}+bx+c=0\] is given as \[\Delta ={{b}^{2}}-4ac\]

Complete step-by-step solution
We are given that the equation of parabola as
\[{{y}^{2}}=4ax....equation(i)\]
We are given that the equation of line that touches the parabola as
\[lx+my+n=0\]
Now, let us find the value of \['x'\] in terms of \['y'\] then we get
\[\begin{align}
  & \Rightarrow lx=-my-n \\
 & \Rightarrow x=-\dfrac{m}{l}y-\dfrac{n}{l} \\
\end{align}\]
We know that if the line touches a parabola that means the line and parabola have only one point of intersection.
We know that we substitute one equation in other equation to find the point of intersection/
Now, by substituting the this value of \['x'\] in equation (i) we get
\[\begin{align}
  & \Rightarrow {{y}^{2}}=4a\left( -\dfrac{m}{l}y-\dfrac{n}{l} \right) \\
 & \Rightarrow {{y}^{2}}+\dfrac{4am}{l}y+\dfrac{4an}{l}=0.......equation(ii) \\
\end{align}\]
Here, we can see that we got the quadratic equation in \['y'\]
We know that if there is a one point of intersection that means the discriminant will be zero.
We know that the discriminant of \[a{{x}^{2}}+bx+c=0\] is given as \[\Delta ={{b}^{2}}-4ac\]
Now, by using the discriminant formula for equation (ii) we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{4am}{l} \right)}^{2}}-4\left( 1 \right)\left( \dfrac{4an}{l} \right)=0 \\
 & \Rightarrow \dfrac{16{{a}^{2}}{{m}^{2}}}{{{l}^{2}}}=\dfrac{16an}{l} \\
\end{align}\]
Now, by cross multiplying the terms in the above equation we get
\[\Rightarrow a{{m}^{2}}=nl\]
Hence the required result has been proved.

Note: We have a direct condition for a tangent of a parabola.
If \[y=Mx+c\] is the tangent to \[{{y}^{2}}=4ax\] then \[c=\dfrac{a}{M}\]
We have the line equation as
\[\begin{align}
  & \Rightarrow lx+my+n=0 \\
 & \Rightarrow y=\left( \dfrac{-l}{m} \right)x+\left( \dfrac{-n}{m} \right) \\
\end{align}\]
Now, by comparing this line equation with the above line equation that is \[y=mx+c\] we get
\[\begin{align}
  & \Rightarrow M=\dfrac{-l}{m} \\
 & \Rightarrow c=\dfrac{-n}{m} \\
\end{align}\]
Now, by using the standard condition that is \[c=\dfrac{a}{M}\] we get
\[\begin{align}
  & \Rightarrow \dfrac{-n}{m}=\dfrac{a}{\left( \dfrac{-l}{m} \right)} \\
 & \Rightarrow \dfrac{nl}{{{m}^{2}}}=a \\
 & \Rightarrow nl=a{{m}^{2}} \\
\end{align}\]
Hence the required result has been proved.