Question

At what point of the parabola ${x^2} = 9y$ is the abscissa three times the ordinate?

Answer 1

Hint: Abscissa i.e. $x - $coordinate is to be three times the ordinate i.e. $y - $coordinate. Assume any arbitrary point $\left( {3a,a} \right)$ and satisfy it with the equation of the parabola.

Complete step-by-step answer:
Let the ordinate i.e. $y - $coordinate of the required point is $a$.
Since, the abscissa is to be three times the ordinate, the abscissa i.e. $x - $coordinate will be $3a$.
Thus, the coordinate of the point is $\left( {3a,a} \right)$. Now, this point is lying on the parabola, so it will satisfy the equation of parabola. And the equation of parabola given in the question is:
$ \Rightarrow {x^2} = 9y$.
Putting $x = 3a$ and $y = a$, we’ll get:
$
   \Rightarrow {\left( {3a} \right)^2} = 9\left( a \right), \\
   \Rightarrow 9{a^2} = 9a, \\
$
$ \Rightarrow a = 0$ or $a = 1$
If we consider $a = 0,$ we have:
$x - $coordinate $ = 0$ and $y - $coordinate $ = 0$.
So, the required point is $\left( {0,0} \right)$
On the other hand if we consider $a = 1,$ we have:
$x - $coordinate $ = 3$ and $y - $coordinate $ = 1$.
The point in this case is $\left( {3,1} \right)$.
Therefore, the points on the parabola for which abscissa three times the ordinate are $\left( {0,0} \right)$ and $\left( {3,1} \right)$.

Note: Whenever a curve is passing through a point, the equation of the curve is always satisfied by that point. This is the condition we used in the above problem.