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

Hint: If the three vertices of a triangle are \[\left( {{a_1},{b_1}} \right),\left( {{a_2},{b_2}} \right),\left( {{a_2},{b_2}} \right)\] then the area of the triangle is given as \[Area = \dfrac{1}{2}\left[ {{a_1}\left( {{a_2} - {a_3}} \right) + {a_2}\left( {{a_3} - {a_1}} \right) + {a_3}\left( {{a_1} - {a_2}} \right)} \right] \]
In this question the ordinated of its vertices is given to us so by assuming the abscissa of the vertices we will get vertices of the triangle and then we will substitute those vertices in the equation of the parabola by which we will find the area of the triangle which is inscribed in the parabola.

Complete step-by-step answer:
Given the equation of the parabola \[{y^2} = 4ax\]
Given \[{y_1},{y_2},{y_3}\] are the ordinated of its vertices of the triangle
Let \[{x_1},{x_2},{x_3}\] be the abscissa of its vertices
So we can write the vertices of the triangle will be \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\]
Now we substitute the vertices of the triangle in the given parabola equation \[{y^2} = 4ax\] , we get
 \[
  y_1^2 = 4a{x_1} \\
  y_2^2 = 4a{x_2} \\
  y_3^2 = 4a{x_3} \\
 \]
Now we will find the area of the triangle for the vertices \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\] as
 \[area\Delta = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right|\]
Now we have already got \[y_1^2 = 4a{x_1},y_2^2 = 4a{x_2},y_3^2 = 4a{x_3}\] , hence we can further write the area as
 \[area\Delta = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {\dfrac{{y_1^2}}{{4a}}}&{{y_1}}&1 \\
  {\dfrac{{y_2^2}}{{4a}}}&{{y_2}}&1 \\
  {\dfrac{{y_3^2}}{{4a}}}&{{y_3}}&1
\end{array}} \right|\]
Take \[\dfrac{1}{{4a}}\] as common from the first column of the matrix, so we can write
 \[
  area\Delta = \dfrac{1}{2} \times \dfrac{1}{{4a}}\left| {\begin{array}{*{20}{c}}
  {y_1^2}&{{y_1}}&1 \\
  {y_2^2}&{{y_2}}&1 \\
  {y_3^2}&{{y_3}}&1
\end{array}} \right| \\
  area\Delta = \dfrac{1}{{8a}}\left| {\begin{array}{*{20}{c}}
  {y_1^2}&{{y_1}}&1 \\
  {y_2^2}&{{y_2}}&1 \\
  {y_3^2}&{{y_3}}&1
\end{array}} \right| \\
 \]
Now by solving the matrix, we get
 \[
\Rightarrow area\Delta = \dfrac{1}{{8a}}\left| {y_1^2\left( {{y_2} - {y_3}} \right) + y_2^2\left( {{y_1} - {y_3}} \right) + y_3^2\left( {{y_1} - {y_2}} \right)} \right| \\
   = \dfrac{1}{{8a}}\left| {y_1^2{y_2} - y_1^2{y_3} + y_2^2{y_1} - y_2^2{y_3} + y_3^2{y_1} - y_3^2{y_2}} \right| \\
 \]
Now by further solving the equation by taking common variables we get’
 \[
\Rightarrow area\Delta = \dfrac{1}{{8a}}\left| {y_1^2{y_2} - y_1^2{y_3} - y_2^2{y_3} - y_3^2{y_2} + y_3^2{y_1} + y_2^2{y_1}} \right| \\
   = \dfrac{1}{{8a}}\left| {y_1^2\left( {{y_2} - {y_3}} \right) - {y_2}{y_3}\left( {{y_2} - {y_3}} \right) - {y_1}\left( {y_3^2 - y_2^2} \right)} \right| \\
   = \dfrac{1}{{8a}}\left| {y_1^2\left( {{y_2} - {y_3}} \right) - {y_2}{y_3}\left( {{y_2} - {y_3}} \right) - {y_1}\left( {{y_3} - {y_2}} \right)\left( {{y_3} + {y_2}} \right)} \right| \\
 \]
Now we take \[\left( {{y_2} - {y_3}} \right)\] as common term, hence we get
 \[area\Delta = \dfrac{1}{{8a}}\left( {{y_2} - {y_3}} \right)\left| {y_1^2 - {y_2}{y_3} - {y_1}\left( {{y_3} + {y_2}} \right)} \right|\]
Now on further solving this determinant equation, we get
 \[
\Rightarrow area\Delta = \dfrac{1}{{8a}}\left( {{y_2} - {y_3}} \right)\left| {y_1^2 - {y_2}{y_3} - {y_1}{y_3} - {y_1}{y_2}} \right| \\
   = \dfrac{1}{{8a}}\left( {{y_2} - {y_3}} \right)\left| {y_1^2 - {y_1}{y_3} - {y_2}{y_3} - {y_1}{y_2}} \right| \\
   = \dfrac{1}{{8a}}\left( {{y_2} - {y_3}} \right)\left| {{y_1}\left( {{y_1} - {y_2}} \right) - {y_3}\left( {{y_1} - {y_2}} \right)} \right| \\
   = \dfrac{1}{{8a}}\left[ {\left( {{y_1} - {y_3}} \right)\left( {{y_2} - {y_3}} \right)\left( {{y_1} - {y_2}} \right)} \right] \\
 \]
Hence proved 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
So, the correct answer is “ \[{y^2} = 4ax\] 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”.

Note: Abscissa is the distance of a point from the y-axis on the graph whereas ordinate is the distance of a point from the x-axis on the graph. In this question ordinate is the distance of a point from the x-axis is given so we will assume three Abscissa on the basis of the ordinate.