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Question: If the area of the triangle inscribed in the parabola \({{y}^{2}}=8x\), then the ordinate of whose v...

If the area of the triangle inscribed in the parabola y2=8x{{y}^{2}}=8x, then the ordinate of whose vertices are 1, 3 and 4 is
A. 34\dfrac{3}{4}
B. 38\dfrac{3}{8}
C. 43\dfrac{4}{3}
D. 54\dfrac{5}{4}

Explanation

Solution

We will be using the concepts of parabola and coordinate geometry to solve the problem. We will be using the equation of the parabola and the y-coordinates given in the question to find the vertices of the triangle then we will use the formulae for finding the area of the triangle in coordinate geometry.

Complete step by step answer:
Now, we have been given a parabola y2=8x{{y}^{2}}=8x. Also, we have been given ordinates of the vertices of triangles inscribed by the parabola. So, we have,
y1=1,y2=3,y3=4{{y}_{1}}=1,{{y}_{2}}=3,{{y}_{3}}=4

Now, since the points lie of y2=8x{{y}^{2}}=8x Therefore, we will use the equation of parabola to find the subsequent x-coordinates. So,
⇒y12=8x1 ⇒12=8x1 ⇒18=x1 ⇒y22=8x2 ⇒98=x2 ⇒y32=8x3 ⇒168=x3 ⇒2=x3 \begin{aligned} & \Rightarrow {{y}_{1}}^{2}=8{{x}_{1}} \\\ & \Rightarrow {{1}^{2}}=8{{x}_{1}} \\\ & \Rightarrow \dfrac{1}{8}={{x}_{1}} \\\ & \Rightarrow {{y}_{2}}^{2}=8{{x}_{2}} \\\ & \Rightarrow \dfrac{9}{8}={{x}_{2}} \\\ & \Rightarrow {{y}_{3}}^{2}=8{{x}_{3}} \\\ & \Rightarrow \dfrac{16}{8}={{x}_{3}} \\\ & \Rightarrow 2={{x}_{3}} \\\ \end{aligned}
So, the vertices of triangle are;
(18,1)(98,3)(2,4)\left( \dfrac{1}{8},1 \right)\left( \dfrac{9}{8},3 \right)\left( 2,4 \right)
Now, we know that area of triangles with (x1,y1)(x2,y2)(x3,y3)\left( {{x}_{1}},{{y}_{1}} \right)\left( {{x}_{2}},{{y}_{2}} \right)\left( {{x}_{3}},{{y}_{3}} \right) as vertices is;
⇒12(x1(y2−y3)+x2(y3−y1)+x3(y1−y2)) Area=12(18(3−4)+98(4−1)+2(1−3)) ⇒12(18(−1)+98(3)+2(−2)) ⇒12(−18+278−4) ⇒12(−1+27−328) ⇒−33+2716 ⇒−616 ⇒−38 \begin{aligned} & \Rightarrow \dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right) \\\ & Area=\dfrac{1}{2}\left( \dfrac{1}{8}\left( 3-4 \right)+\dfrac{9}{8}\left( 4-1 \right)+2\left( 1-3 \right) \right) \\\ & \Rightarrow \dfrac{1}{2}\left( \dfrac{1}{8}\left( -1 \right)+\dfrac{9}{8}\left( 3 \right)+2\left( -2 \right) \right) \\\ & \Rightarrow \dfrac{1}{2}\left( \dfrac{-1}{8}+\dfrac{27}{8}-4 \right) \\\ & \Rightarrow \dfrac{1}{2}\left( \dfrac{-1+27-32}{8} \right) \\\ & \Rightarrow \dfrac{-33+27}{16} \\\ & \Rightarrow \dfrac{-6}{16} \\\ & \Rightarrow \dfrac{-3}{8} \\\ \end{aligned}
Since, the area can’t be negative therefore the area is 38\dfrac{3}{8} .

So, the correct answer is “Option B”.

Note: To solve these type of question it is important to remember the concepts of parabola and coordinate geometry also the question can be solved alternatively as;

& Area=\dfrac{\left( {{y}_{1}}-{{y}_{2}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left( {{y}_{3}}-{{y}_{1}} \right)}{8a} \\\ & \Rightarrow \dfrac{\left( 1-3 \right)\left( 3-4 \right)\left( 4-1 \right)}{8\times 2} \\\ & \Rightarrow \dfrac{3}{8} \\\ \end{aligned}$$