Question

Find the locus of the middle points of the chords of parabola${{y}^{2}}=8x$ if slope of the chords be 4?
a) $y=2$
b) $y+x=1$
c) $y=1$
d) $y=0$

Answer 1

Hint: Consider the points on parabola in the form of$(a{{t}^{2}},2at)$ and find the equation of line using the two points taken on parabola.


Complete step-by-step solution -

Let the given parabola ${{y}^{2}}=4ax$ has a chord, which cuts parabola at $(a{{t}_{1}}^{2},2a{{t}_{1}})$& $(a{{t}_{2}}^{2},2a{{t}_{2}})$ .

Let the given midpoint of chord AB be$\left( h,k \right)$ .
We know midpoint of a line joining two point $\left( x,y \right)\And \left( m,n \right)$ =$\left( \dfrac{x+m}{2},\dfrac{y+n}{2} \right)$ .
$h=\left( \dfrac{a{{t}^{2}}_{1}+a{{t}^{2}}_{2}}{2} \right)$ & $k=\left( \dfrac{2a{{t}_{1}}+2a{{t}_{2}}}{2} \right)$
$h=\dfrac{a}{2}\left( {{t}^{2}}_{1}+{{t}^{2}}_{2} \right)$ …(1)
& $k=a\left( {{t}_{1}}+{{t}_{2}} \right)$ …(2)
Now,
$2h=a\left( {{t}^{2}}_{1}+{{t}^{2}}_{2} \right)$
Now manipulating it as ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ ,
$2h=a{{\left( ({{t}_{1}}+{{t}_{2}} \right)}^{2}}-2{{t}_{1}}{{t}_{2}})$
From equation (2).
$\begin{align}
  & 2h=a\left( {{\left( \dfrac{k}{a} \right)}^{2}}-2{{t}_{1}}{{t}_{2}} \right) \\
 & 2h=a\left( {{\dfrac{k}{{{a}^{2}}}}^{2}}-2{{t}_{1}}{{t}_{2}} \right) \\
\end{align}$
${{t}_{1}}{{t}_{2}}=\dfrac{{{k}^{2}}}{2{{a}^{2}}}-\dfrac{h}{a}$ …(3)
Now find the equation of chord passing through two point at $(a{{t}_{1}}^{2},2a{{t}_{1}})$& $(a{{t}_{2}}^{2},2a{{t}_{2}})$ .
Using two point formula of equation of line $y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{1}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$ .
Then,
$y-2a{{t}_{1}}=\dfrac{\left( 2a{{t}_{2}}-2a{{t}_{1}} \right)}{a{{t}^{2}}_{2}-a{{t}^{2}}_{1}}\left( x-a{{t}^{2}}_{1} \right)$
Taking a common from numerator and denominator & then expand denominator $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ .
$\begin{align}
  & y-2a{{t}_{1}}=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{{{t}^{2}}_{2}-{{t}^{2}}_{1}}\left( x-a{{t}^{2}}_{1} \right) \\
 & y-2a{{t}_{1}}=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{({{t}_{2}}+{{t}_{1}})\left( {{t}_{2}}-{{t}_{1}} \right)}\left( x-a{{t}^{2}}_{1} \right) \\
 & y-2a{{t}_{1}}=\dfrac{2}{({{t}_{2}}+{{t}_{1}})}\left( x-a{{t}^{2}}_{1} \right) \\
 & (y-2a{{t}_{1}})({{t}_{2}}+{{t}_{1}})=2\left( x-a{{t}^{2}}_{1} \right) \\
 & y({{t}_{2}}+{{t}_{1}})-2a{{t}_{1}}{{t}_{2}}-2a{{t}^{2}}_{1}=2x-2a{{t}^{2}}_{1} \\
\end{align}$
$-2a{{t}^{2}}_{1}$ will be cancel out from both side
Then, we get
$\begin{align}
  & y({{t}_{2}}+{{t}_{1}})-2a{{t}_{1}}{{t}_{2}}=2x \\
 & \\
\end{align}$
Equation of chord will be:
$y=\dfrac{2x}{({{t}_{2}}+{{t}_{1}})}+\dfrac{2a{{t}_{1}}{{t}_{2}}}{({{t}_{2}}+{{t}_{1}})}$ …(4)
Now,
Parabola equation given: ${{y}^{2}}=8x$
Comparing with the standard equation, we get $a=2$ .
Then equation of chord after putting value of a:
$y=\dfrac{2x}{({{t}_{2}}+{{t}_{1}})}+\dfrac{4{{t}_{1}}{{t}_{2}}}{({{t}_{2}}+{{t}_{1}})}$
Slope of chord $=4$ (given in the question)
$\dfrac{2}{({{t}_{2}}+{{t}_{1}})}=4$ (from equation (4) comparing with the general equation of line $y=mx+c$ )
From equation (2)
$\begin{align}
  & \dfrac{2}{\dfrac{k}{2}}=2 \\
 & \dfrac{2}{k}=2 \\
 & k=1 \\
\end{align}$
Replace k with $y$, we get
$y=1$
Hence, option c) is true.

Note: We can use direct formula to find the equation of tangent taking two arbitrary point on a parabola by using direct formula $y({{t}_{2}}+{{t}_{1}})-2a{{t}_{1}}{{t}_{2}}=2x$. In this question writing the equation of tangent in the form of $y=mx+c$ as slope $'m'$ is given in the question.