Question

Trace the parabola: ${{\left( 5x-12y \right)}^{2}}=2ax+29ay+{{a}^{2}}$

Answer 1

We will introduce the variable λ in the given equation. We will add the variable λ on the LHS side of the equation. To balance the equation on both sides we have to add the same variables as in the LHS. We will suppose that LHS and RHS are equal to zero and will get two equations of a straight line. We will assume that these two straight lines are perpendicular to each other. We know that if two lines are perpendicular, then $mm'=-1$ .We can calculate the value of using this approach. We will now substitute the value of λ. As we know the general equation of parabola is ${{y}^{2}}=4ax$ we will try to arrange the expression to get this form.

Complete step-by-step answer:
We have equation ${{\left( 5x-12y \right)}^{2}}=2ax+29ay+{{a}^{2}}$
We will now add variable λ on the LHS of equation and we will balance the equation by addin required terms, we get,
$\Rightarrow {{\left( 5x-12y+\lambda \right)}^{2}}=2ax+29ay+{{a}^{2}}-24\lambda y+10\lambda x+{{\lambda }^{2}}$
We write above equation as,
$\Rightarrow {{\left( 5x-12y+\lambda \right)}^{2}}=\left( 2a+10\lambda \right)x+\left( 29a-24\lambda \right)y+{{a}^{2}}+{{\lambda }^{2}}$
We will suppose LHS=RHS=0, we will get
$\Rightarrow 5x-12y+\lambda =0.......(i)$
$\Rightarrow \left( 2a+10\lambda \right)x+\left( 29a-24\lambda \right)y+{{a}^{2}}+{{\lambda }^{2}}=0.......(ii)$
Let assume that the equation(i) and equation(ii) are two straight line that are perpendicular, we will get,
$\Rightarrow {{m}_{1}}{{m}_{2}}=-1$
We can write equation(i) and equation(ii) in form $y=mx+c$ where m is slope of line,
$\Rightarrow y=\dfrac{5}{12}x+\dfrac{\lambda }{12}$
We get ${{m}_{1}}=\dfrac{5}{12}$
$\Rightarrow y=-\dfrac{\left( 2a+10\lambda \right)}{\left( 29a-24\lambda \right)}x-\dfrac{{{a}^{2}}+{{\lambda }^{2}}}{\left( 29a-24\lambda \right)}$
We get ${{m}_{2}}=-\dfrac{\left( 2a+10\lambda \right)}{\left( 29a-24\lambda \right)}$
We know that, ${{m}_{1}}{{m}_{2}}=-1$,
$\Rightarrow \dfrac{5}{12}\times -\dfrac{\left( 2a+10\lambda \right)}{\left( 29a-24\lambda \right)}=-1$
We will solve above expression, we get,
$\begin{aligned} & \Rightarrow 10a+50\lambda =348a-288\lambda \\\ & \Rightarrow 338\lambda =338a \\\ & \Rightarrow \lambda =a \\\ \end{aligned}$
We will substitute this in equation, we will get,
$\Rightarrow {{\left( 5x-12y+\lambda \right)}^{2}}=\left( 2a+10\lambda \right)x+\left( 29a-24\lambda \right)y+{{a}^{2}}+{{\lambda }^{2}}$
$\begin{aligned} & \Rightarrow {{\left( 5x-12y+a \right)}^{2}}=\left( 2a+10a \right)x+\left( 29a-24a \right)y+{{a}^{2}}+{{a}^{2}} \\\ & \Rightarrow {{\left( 5x-12y+a \right)}^{2}}=12ax+5ay+2{{a}^{2}} \\\ \end{aligned}$
We will take a common from RHS,
$\Rightarrow {{\left( 5x-12y+a \right)}^{2}}=a\left( 12x+5y+2a \right)$
We know that general form of equation of parabola is,
$\Rightarrow {{Y}^{2}}=4a'{{X}^{2}}$
Comparing both the equation we get,
$\begin{aligned} & \Rightarrow Y=5x-12y+a \\\ & \Rightarrow X=12x+5y+2a \\\ & \Rightarrow 4a'=a \\\ \end{aligned}$
So, the equation of parabola is
$\Rightarrow {{Y}^{2}}=\dfrac{a}{4}X$
Where $Y=5x-12y+a$ and $X=12x+5y+2a$

Note: There will be some questions where we have to assume few things from our side to solve the question like in this question, we have introduced variable λ. To solve the questions of conic sections it is important that the student is thorough with the concepts of straight lines.