Question

Prove that the area of the triangle inscribed in the parabola ${{y}^{2}}=4ax$ is $\dfrac{1}{8a}\left| \left( {{y}_{1}}-{{y}_{2}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left( {{y}_{3}}-{{y}_{1}} \right) \right|$ sq. units where ${{y}_{1}},{{y}_{2}},{{y}_{3}}$ are the ordinates of its vertices.

Answer 1

We consider the 3 points of the coordinates of the triangle as $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right).$ We take the x-coordinates in terms of ${{y}_{1}},{{y}_{2}},{{y}_{3}}$ only. Since the triangle is inscribed in the parabola, we know that the vertices of the triangle lie on the parabola. Then using these coordinates, we calculate the area of the triangle using the formula $Area=\dfrac{1}{2}\left| D \right|,$ where D stands for the determinant of the matrix for the coordinates of the triangle.

Complete step by step solution:
In order to solve this question, let us consider the coordinates of the three vertices of the triangle to be $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right).$ Now we know that since the triangle is inscribed in the parabola, the vertices of the triangle lie on the parabola. We can see this in the image below.

Hence, the coordinate points of the triangle satisfy the equation of the parabola as ${{y}_{1}}^{2}=4a{{x}_{1}},{{y}_{2}}^{2}=4a{{x}_{2}},{{y}_{3}}^{2}=4a{{x}_{3}}.$ Rearranging this in terms of ${{y}_{1}},{{y}_{2}},{{y}_{3}}$ , we get the coordinates completely in terms of ${{y}_{1}},{{y}_{2}},{{y}_{3}}$ as $\left( \dfrac{{{y}_{1}}^{2}}{4a},{{y}_{1}} \right),\left( \dfrac{{{y}_{2}}^{2}}{4a},{{y}_{2}} \right),\left( \dfrac{{{y}_{3}}^{2}}{4a},{{y}_{3}} \right).$
We now calculate the area of the triangle using the formula,
$\Rightarrow Area=\dfrac{1}{2}\left| D \right|$
Where $\left| D \right|$ is the determinant of the coordinates of the triangle given by
$\Rightarrow \left| D \right|=\left| \begin{matrix} \dfrac{{{y}_{1}}^{2}}{4a} & {{y}_{1}} & 1 \\\ \dfrac{{{y}_{2}}^{2}}{4a} & {{y}_{2}} & 1 \\\ \dfrac{{{y}_{3}}^{2}}{4a} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$
Taking the term $\dfrac{1}{4a}$ common from the first column of the determinant,
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left| \begin{matrix} {{y}_{1}}^{2} & {{y}_{1}} & 1 \\\ {{y}_{2}}^{2} & {{y}_{2}} & 1 \\\ {{y}_{3}}^{2} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$
Using row operations, we subtract the rows 1 and 2 with row 3,
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left| \begin{matrix} {{y}_{1}}^{2}-{{y}_{3}}^{2} & {{y}_{1}}-{{y}_{3}} & 1-1 \\\ {{y}_{2}}^{2}-{{y}_{3}}^{2} & {{y}_{2}}-{{y}_{3}} & 1-1 \\\ {{y}_{3}}^{2} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$
Splitting the ${{y}_{1}}^{2}-{{y}_{3}}^{2}$ as $\left( {{y}_{1}}-{{y}_{3}} \right)\left( {{y}_{1}}+{{y}_{3}} \right)$ and ${{y}_{2}}^{2}-{{y}_{3}}^{2}$ as $\left( {{y}_{2}}-{{y}_{3}} \right)\left( {{y}_{2}}+{{y}_{3}} \right),$
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left| \begin{matrix} \left( {{y}_{1}}-{{y}_{3}} \right)\left( {{y}_{1}}+{{y}_{3}} \right) & {{y}_{1}}-{{y}_{3}} & 0 \\\ \left( {{y}_{2}}-{{y}_{3}} \right)\left( {{y}_{2}}+{{y}_{3}} \right) & {{y}_{2}}-{{y}_{3}} & 0 \\\ {{y}_{3}}^{2} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$
Taking the terms $\left( {{y}_{1}}-{{y}_{3}} \right)$ common out from the first row and the term $\left( {{y}_{2}}-{{y}_{3}} \right)$ common from the second row,
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left( {{y}_{1}}-{{y}_{3}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left| \begin{matrix} \left( {{y}_{1}}+{{y}_{3}} \right) & 1 & 0 \\\ \left( {{y}_{2}}+{{y}_{3}} \right) & 1 & 0 \\\ {{y}_{3}}^{2} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$
Taking the determinant,
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left( {{y}_{1}}-{{y}_{3}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left[ \left( {{y}_{1}}+{{y}_{3}} \right).\left( 1\times 1-{{y}_{3}}\times 0 \right)-1.\left( \left( {{y}_{2}}+{{y}_{3}} \right)\times 1-{{y}_{3}}^{2}\times 0 \right)+0.\left( \left( {{y}_{2}}+{{y}_{3}} \right)\times {{y}_{3}}-{{y}_{3}}^{2}\times 1 \right) \right]$
Taking the product of the first two terms and since the product of the third term with 0 is 0, we ignore the last term,
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left( {{y}_{1}}-{{y}_{3}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left[ \left( {{y}_{1}}+{{y}_{3}} \right)\times \left( 1-0 \right)-\left( \left( {{y}_{2}}+{{y}_{3}} \right)-0 \right) \right]$
Multiplying the first term with 1 and writing the second term as it is,
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left( {{y}_{1}}-{{y}_{3}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left[ \left( {{y}_{1}}+{{y}_{3}} \right)-\left( {{y}_{2}}+{{y}_{3}} \right) \right]$
Taking the second term outside the bracket and multiplying with the negative sign,
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left( {{y}_{1}}-{{y}_{3}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left[ {{y}_{1}}+{{y}_{3}}-{{y}_{2}}-{{y}_{3}} \right]$
Subtracting the ${{y}_{3}}$ terms,
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left( {{y}_{1}}-{{y}_{3}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left[ {{y}_{1}}-{{y}_{2}} \right]$
Taking the magnitude since are is always positive, we can rearrange the terms in the first bracket as,
$\Rightarrow \left| D \right|=\dfrac{1}{4a}\left| \left( {{y}_{3}}-{{y}_{1}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left( {{y}_{1}}-{{y}_{2}} \right) \right|$
Substituting this whole term in the formula for area, we get
$\Rightarrow Area=\dfrac{1}{2}\times \dfrac{1}{4a}\left| \left( {{y}_{1}}-{{y}_{2}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left( {{y}_{3}}-{{y}_{1}} \right) \right|$
Multiplying the constants,
$\Rightarrow Area=\dfrac{1}{8a}\left| \left( {{y}_{1}}-{{y}_{2}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left( {{y}_{3}}-{{y}_{1}} \right) \right|$
Hence, we have proved that the area of the triangle is $\dfrac{1}{8a}\left| \left( {{y}_{1}}-{{y}_{2}} \right)\left( {{y}_{2}}-{{y}_{3}} \right)\left( {{y}_{3}}-{{y}_{1}} \right) \right|.$

Note:
We need to know the basic concepts of finding the determinant in order to solve this question easily. We need to be careful while solving the determinant. It becomes easier to first do the row operations and manipulations before calculating the determinant than to directly calculate the determinant. We can also directly solve the determinant but it can be time consuming and more error prone.