Question

Trace the parabolas \[144{{x}^{2}}-120xy+25{{y}^{2}}+619x-272y+663=0\], and find its focus.

Answer 1

Hint:In this question, we first need to add and rearrange the given equation with suitable terms. Then convert it to the standard form and by using the formula for the focus in a standard form we can get the result.

Complete step-by-step answer:
From the given equation in the question we have,
 \[\Rightarrow 144{{x}^{2}}-120xy+25{{y}^{2}}+619x-272y+663=0\]
In order to compare the given equation with the standard form of parabola we first need to express it in terms of square
As we already know that standard form of a parabola is expressed as
\[{{y}^{2}}=4ax\]
Now, let us rearrange the terms and write them as a square term on one side.
\[\Rightarrow {{\left( 12x-5y \right)}^{2}}=272y-619x-663\]
Here, as we need to express the given equation in standard form we need to introduce a constant such that on simplification we can obtain the standard form accordingly
Let us now introduce a \[\lambda \] term on the left hand side and subtract those respective \[\lambda \] terms on the right hand side.
\[\Rightarrow {{\left( 12x-5y+\lambda \right)}^{2}}=272y-619x-663-10\lambda y+24\lambda x+{{\lambda }^{2}}\]
Let us now write the x and y terms together on the right hand side then we get,
\[\Rightarrow {{\left( 12x-5y+\lambda \right)}^{2}}=\left( 272-10\lambda \right)y-\left( 619-24\lambda \right)x+{{\lambda }^{2}}-663\]
Now, on equating the left hand side and right hand side of the above equation separately to 0 we get,
\[\begin{align}
  & \Rightarrow 12x-5y+\lambda =0.......\left( 1 \right) \\
 & \Rightarrow \left( 272-10\lambda \right)y-\left( 619-24\lambda \right)x+{{\lambda }^{2}}-663=0 \\
\end{align}\]
For the value of \[\lambda \] we assume that both the lines are perpendicular.
As we already know that when two lines are perpendicular then the product of their slopes is -1.
\[m\times m'=-1\]
Let us assume the slope of the line in equation (1) as m and the slope of the line in equation (2) as m'
Now, the slopes of the lines are as follows:
 \[\begin{align}
  & \Rightarrow m=-\dfrac{12}{-5} \\
 & \therefore m=\dfrac{12}{5} \\
 & \Rightarrow m'=-\dfrac{-\left( 619-24\lambda \right)}{\left( 272-10\lambda \right)} \\
 & \therefore m'=\dfrac{\left( 619-24\lambda \right)}{\left( 272-10\lambda \right)} \\
\end{align}\]
Now, by using the condition for perpendicular of the lines we get,
\[\Rightarrow m\times m'=-1\]
Now, on substituting the respective values we get,
\[\Rightarrow \dfrac{12}{5}\times \dfrac{\left( 619-24\lambda \right)}{\left( 272-10\lambda \right)}=-1\]
Let us now cross multiply the terms and rearrange them.
\[\Rightarrow 7428-288\lambda =50\lambda -1360\]
\[\Rightarrow 338\lambda =8788\]
Now, by dividing with 338 on both sides we get,
\[\therefore \lambda =26\]
Now, by substituting this value of \[\lambda \] back we get,
\[\Rightarrow {{\left( 12x-5y+26 \right)}^{2}}=\left( 272-10\times 26 \right)y-\left( 619-24\times 26 \right)x+{{\left( 26 \right)}^{2}}-663\]
Now, on further simplification of the terms in the above equation we get,
\[\begin{align}
  & \Rightarrow {{\left( 12x-5y+26 \right)}^{2}}=\left( 272-260 \right)y-\left( 619-624 \right)x+676-663 \\
 & \Rightarrow {{\left( 12x-5y+26 \right)}^{2}}=5x+12y+13 \\
\end{align}\]
Now, by dividing it with square of 13 on both sides we get,
\[\Rightarrow {{\left( \dfrac{12x-5y+26}{13} \right)}^{2}}=\dfrac{1}{13}\left( \dfrac{5x+12y+13}{13} \right)\]
Now let us assume the term on the left hand side as Y and term on the right hand side as X as follows:
\[\dfrac{12x-5y+26}{13}=Y,\dfrac{5x+12y+13}{13}=X\]
Now, the equations changes as
\[\Rightarrow {{Y}^{2}}=\dfrac{1}{13}X\]
Now, on comparing this with the standard form of the parabola \[{{y}^{2}}=4ax\] we get,
\[\begin{align}
  & \Rightarrow 4a=\dfrac{1}{13} \\
 & \therefore a=\dfrac{1}{52} \\
\end{align}\]
Now, the focus of the parabola can be obtained by solving the equations
\[X-a=0,Y=0\]
Now, the respective equations are
\[\Rightarrow \dfrac{5x+12y+13}{13}-\dfrac{1}{52}=0\]
Now, on rearranging and cross multiplying we get,
\[\Rightarrow 20x+48y+51=0....\left( 3 \right)\]
\[\begin{align}
  & \Rightarrow \dfrac{12x-5y+26}{13}=0 \\
 & \Rightarrow 12x-5y+26=0......\left( 4 \right) \\
\end{align}\]
Now, on solving the equations (3) and (4) we get,
\[\therefore x=\dfrac{-1503}{676},y=\dfrac{-23}{169}\]
Hence, the focus of the parabola is \[\left( \dfrac{-1503}{676},\dfrac{-23}{169} \right)\].

Note:It is important to note that as the given equation is not in the standard form we need to convert it to the standard form to get the focus. So, we need to assume some variable and introduce it to the given equation then solve according to get the value of the variable.
Here, we cannot directly get the focus because in the standard form equation we have equations in terms of x and y. As focus of the parabola satisfies the equation of the parabola so on solving those equations obtained gives us the coordinates of the focus.