Question

Find the length of the latus rectum of the parabola whose parametric equation are given by \[x={{t}^{2}}+t+1\text{ and }y={{t}^{2}}-t+1\]

Answer 1

Hint: We will first solve for the value of \[t\] using both the equations. Then put the value of \[t\] in any one equation. We will get the equation of the parabola. We know that the point on the parabola is equidistant from a fixed line and a fixed point. We will write the equation in Focus- Directrix form, given as \[{{\left( \dfrac{x+y+k}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right)}^{2}}={{\left( x-a \right)}^{2}}+{{\left( x-a \right)}^{2}}\] . We will solve the equation to get the value of variables. We know that the distance between Directrix and focus is ‘2a’ and the line joining focus and directrix is perpendicular to directrix. We will now calculate the distance. We also know that the length of the latus rectum is equal to ‘4a’

Complete step-by-step answer:
We have these two equations
\[\begin{align}
  & \Rightarrow x={{t}^{2}}+t+1.....(i) \\
 & \Rightarrow y={{t}^{2}}-t+1......(ii) \\
\end{align}\]
Adding EQ.( \[i\]) And Eq.( \[ii\])
\[\begin{align}
  & \Rightarrow x+y=2\left( {{t}^{2}}+1 \right) \\
 & \Rightarrow x+y-2=2{{t}^{2}}.....(iii) \\
\end{align}\]
We will now subtract equation(ii) from equation (i)
\[\Rightarrow x-y=2t......(iv)\]
We will solve equation (iii) and equation (iv)
\[\begin{align}
  & \Rightarrow x+y-2=2\left( \dfrac{{{\left( x-y \right)}^{2}}}{4} \right) \\
 & \Rightarrow x+y-2=\dfrac{{{\left( x-y \right)}^{2}}}{2} \\
 & \Rightarrow 2(x+y-2)={{\left( x-y \right)}^{2}} \\
 & \Rightarrow 2x+2y-4={{x}^{2}}+{{y}^{2}}-2xy \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-2xy-2x-2y+4=0.......(v) \\
 & \\
\end{align}\]
 The above equation can be written in the focus-directrix form as
\[{{\left( \dfrac{x+y+k}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right)}^{2}}={{\left( x-a \right)}^{2}}+{{\left( x-a \right)}^{2}}\]
Where \[x+y+k=0\] is the equation of directrix and (a, a) is the focus.
The values of a and k can be found by equating the coefficient of the original equation(v).
Equating the constant term, we get,
\[\begin{align}
  & 4{{a}^{2}}-{{k}^{2}}=4......(vi) \\
 & \\
 & \text{Equating the coefficient of }x\text{, we get,} \\
 & \Rightarrow -4a-2k=-2......(vii) \\
 & \\
 & \text{From }(vi)\text{ and }(vii)\text{ we get }a=\dfrac{5}{4}\text{ and }k=-\dfrac{3}{2} \\
\end{align}\]
So, the equation of directrix is \[x+y-\dfrac{3}{2}=0\]and the coordinates of the focus are \[\left( \dfrac{5}{4},\dfrac{5}{4} \right)\]
We know the perpendicular distance between a line and a point is given by
\[d=\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|\]
Distance between focus and directrix is 2a
\[\begin{align}
  & 2a=\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right| \\
 & \\
 & \text{where }A=1,\text{ }B=1\text{ and }C=-\dfrac{3}{2} \\
 & \text{and }{{x}_{1}}=\dfrac{5}{4}\text{ and }{{y}_{1}}=\dfrac{5}{4} \\
 & \Rightarrow 2a\text{=}\left| \dfrac{\dfrac{5}{4}+\dfrac{5}{4}-\dfrac{3}{2}}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right| \\
 & \Rightarrow 2a\text{=}\dfrac{1}{\sqrt{2}} \\
\end{align}\]
We know that the length of the latus rectum is 4a
Length of the latus rectum \[=2\times \dfrac{1}{\sqrt{2}}=\sqrt{2}\]

Note: Alternate Method:
We have these two equations,
\[\begin{align}
  & \Rightarrow x={{t}^{2}}+t+1.....(i) \\
 & \Rightarrow y={{t}^{2}}-t+1......(ii) \\
\end{align}\]
 Subtracting (i) and (ii), we get
\[\Rightarrow t=\dfrac{\left( x-y \right)}{2}\]
Substituting the value of t in (i)
\[\begin{align}
  & \Rightarrow x={{\left( \dfrac{x-y}{2} \right)}^{2}}+\left( \dfrac{x-y}{2} \right)+1 \\
 & \\
\end{align}\]
Arranging above equation, we get
\[\begin{align}
  & \Rightarrow {{\left( x-y \right)}^{2}}=2\left( x+y-2 \right) \\
 & \\
\end{align}\]
We can write above equation as
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{x-y}{\sqrt{2}} \right)}^{2}}=\left( x+y-2 \right) \\
 & \\
\end{align}\]
We will now multiply and divide LHS by \[\sqrt{2}\]
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{x-y}{\sqrt{2}} \right)}^{2}}=\sqrt{2}\left( \dfrac{x+y-2}{\sqrt{2}} \right) \\
 & \\
\end{align}\]
Equating the above equation with the general form \[{{y}^{2}}=4ax\], we get,
\[\Rightarrow \text{4}a=\sqrt{2}\]
We know that latus rectum of parabola is equal to 4a
\[\therefore \] latus rectum of parabola is \[\sqrt{2}\]