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.

& 4y=3x+8 \\\ & \Rightarrow 4y-8=3x \\\ & \therefore x=\dfrac{4y-8}{3}.........(1) \\\ \end{aligned}$$ Now let us put the value of x in the equation of parabola. $${{y}^{2}}=8x\Rightarrow {{y}^{2}}=8\left( \dfrac{4y-8}{3} \right)$$ Cross multiply the above equation and simplify it. $$\begin{aligned} & {{y}^{2}}=8\left( \dfrac{4y-8}{3} \right) \\\ & \Rightarrow 3{{y}^{2}}=32y-64 \\\ & 3{{y}^{2}}-32y+64=0.......(2) \\\ \end{aligned}$$ The above equation is in the form of a quadratic equation, $$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{aligned} & 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{aligned}$$ Thus we get, $$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{aligned} & =\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{aligned}$$ Thus we got the length of the chord as $$\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.