Question

Find the length of the chord \[4y=3x+8\] intercepted by the parabola \[{{y}^{2}}=8x\].

Answer 1

Hint: Any chord of a parabola will intersect it at two points , find these two points by substituting the value of $x$ or $y$ from the equation of chord in the equation of parabola . Use distance between two point formula to get the answer.

Complete step-by-step answer:
We have been given the equation of a line as \[4y=3x+8\] and we got the equation of the parabola as \[{{y}^{2}}=8x\].
The ordinates of point of intersection of the line \[4y=3x+8\] and parabola \[{{y}^{2}}=8x\] are w.r.t to the equation formed.
\[\begin{align}
  & 4y=3x+8 \\
 & \Rightarrow 4y-8=3x \\
 & \therefore x=\dfrac{4y-8}{3}.........(1) \\
\end{align}\]
Now let us put the value of x in the equation of parabola.
\[{{y}^{2}}=8x\Rightarrow {{y}^{2}}=8\left( \dfrac{4y-8}{3} \right)\]
Cross multiply the above equation and simplify it.
\[\begin{align}
  & {{y}^{2}}=8\left( \dfrac{4y-8}{3} \right) \\
 & \Rightarrow 3{{y}^{2}}=32y-64 \\
 & 3{{y}^{2}}-32y+64=0.......(2) \\
\end{align}\]
The above equation is in the form of a quadratic equation, \[a{{y}^{2}}+by+c=0\].
Thus let us compare both the equations and get the value.
a = 3, b = -32, c = 64.
Now let us substitute these values in the quadratic formula,
\[\begin{align}
  & y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & y=\dfrac{-(-32)\pm \sqrt{{{(-32)}^{2}}-4\times 3\times 64}}{2\times 3}=\dfrac{32\pm \sqrt{1024-768}}{6} \\
 & =\dfrac{32\pm \sqrt{256}}{6}=\dfrac{32\pm 16}{6} \\
\end{align}\]
Thus we get, \[y=\dfrac{32+16}{6}=\dfrac{48}{6}=8\] and \[y=\dfrac{32-16}{6}=\dfrac{16}{6}=\dfrac{8}{3}\].
Thus we got y = 8 and y = \[\dfrac{8}{3}\].
Let us substitute the value in equation (1) and get x.
When y = 8, \[x=\dfrac{4\times 8-8}{3}=\dfrac{32-8}{3}=\dfrac{24}{3}=8\].
Thus when y = 8, then x = 8.
When y = \[\dfrac{8}{3}\], \[x=\dfrac{4\times {}^{8}/{}_{3}-8}{3}=\dfrac{4\times 8-8\times 3}{3\times 3}=\dfrac{8}{9}\].
Thus when y = \[\dfrac{8}{3}\], then x = \[\dfrac{8}{9}\].
Thus we got two points as (8, 8) and \[\left( \dfrac{8}{9},\dfrac{8}{3} \right)\]. Let us consider these points as P \[\left( \dfrac{8}{9},\dfrac{8}{3} \right)\] and Q (8, 8).
The line \[4y=3x+8\] and parabola- \[{{y}^{2}}=8x\] are at P \[\left( \dfrac{8}{9},\dfrac{8}{3} \right)\] and Q (8, 8).
The length of chord PQ can be given by the distance formula,
\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}=\] length of chord.
\[\left( {{x}_{1}},{{y}_{1}} \right)=P\left( \dfrac{8}{9},\dfrac{8}{3} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=Q\left( 8,8 \right)\]
\[\therefore \]Length of the chord PQ,
\[\begin{align}
  & =\sqrt{{{\left( 8-{}^{8}/{}_{9} \right)}^{2}}+{{\left( 8-{}^{8}/{}_{3} \right)}^{2}}} \\
 & =\sqrt{{{\left( \dfrac{72-8}{9} \right)}^{2}}+{{\left( \dfrac{24-8}{3} \right)}^{2}}} \\
 & =\sqrt{{{\left( \dfrac{64}{9} \right)}^{2}}+{{\left( \dfrac{16}{3} \right)}^{2}}} \\
 & =\sqrt{{{\left( \dfrac{4096}{81} \right)}^{2}}+{{\left( \dfrac{256}{9} \right)}^{2}}} \\
 & =\sqrt{\dfrac{4096+2304}{81}} \\
 & =\sqrt{\dfrac{6400}{81}}=\dfrac{80}{9} \\
\end{align}\]
Thus we got the length of the chord as \[\dfrac{80}{9}\].

Note: You can only simplify it by getting the value of x and substituting in the equation of parabola. It is important that you know the basics to solve the quadratic equation and the formula to calculate distance between two points.