Question

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

Answer 1

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

& \Rightarrow lx=-my-n \\\ & \Rightarrow x=-\dfrac{m}{l}y-\dfrac{n}{l} \\\ \end{aligned}$$ We know that if the line touches a parabola that means the line and parabola have only one point of intersection. We know that we substitute one equation in other equation to find the point of intersection/ Now, by substituting the this value of $$'x'$$ in equation (i) we get $$\begin{aligned} & \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{aligned}$$ Here, we can see that we got the quadratic equation in $$'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{aligned} & \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{aligned}$$ Now, by cross multiplying the terms in the above equation we get $$\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{aligned} & \Rightarrow lx+my+n=0 \\\ & \Rightarrow y=\left( \dfrac{-l}{m} \right)x+\left( \dfrac{-n}{m} \right) \\\ \end{aligned}$$ Now, by comparing this line equation with the above line equation that is $$y=mx+c$$ we get $$\begin{aligned} & \Rightarrow M=\dfrac{-l}{m} \\\ & \Rightarrow c=\dfrac{-n}{m} \\\ \end{aligned}$$ Now, by using the standard condition that is $$c=\dfrac{a}{M}$$ we get $$\begin{aligned} & \Rightarrow \dfrac{-n}{m}=\dfrac{a}{\left( \dfrac{-l}{m} \right)} \\\ & \Rightarrow \dfrac{nl}{{{m}^{2}}}=a \\\ & \Rightarrow nl=a{{m}^{2}} \\\ \end{aligned}$$ Hence the required result has been proved.