Question

Find the equation of the parabola which is symmetric about the y-axis and passes through the point (3, -9).

Answer 1

Hint: We solve this by finding where the points (3, -9) lie in the equation of parabola. Since the parabola is symmetric about the y-axis is $$x^2=4ay$$ or $$x^2=-4ay$$ . By using the given points (3, -9) we can find the value of ‘a’ and substituting in the required equation we will get the equation of parabola.

Complete step-by-step answer:
We know the equation of parabola about the y-axis is $$x^2=4ay$$ or $$x^2=-4ay$$ .
Since the points (3, -9) lie in the fourth quadrant so the equation of parabola about the y-axis is $$x^2=-4ay$$ . See the below diagram you will understand easily.

Hence, the equation of parabola is $$x^2=-4ay$$ . ---- (1)
We need to find the value of ‘a’.
Now parabola passes through (3, -9), put $$x=3$$ and $$y=-9$$ in equation (1). We get:
 $$\Rightarrow 3^2=-4\times a\times (-9)$$
 $$\Rightarrow 9=-4\times a\times (-9)$$
We know the product of negative and negative is positive and rearranging the above equation we get:
 $$\Rightarrow a={\displaystyle \frac{9}{9\times 4}}$$
 $$\Rightarrow a={\displaystyle \frac{1}{4}}$$
We know the value of ‘a’. Now substituting in the equation (1). We get
 $$\Rightarrow x^2=-4\left({\displaystyle \frac{1}{4}}\right)y$$
Cancelling 4,
 $$\Rightarrow x^2=-y$$
 $$\Rightarrow x^2+y=0$$ Is the required equation.
The equation of the parabola which is symmetric about the y-axis and passes through the point (3, -9) is $$x^2+y=0$$ .
So, the correct answer is “ $$x^2+y=0$$ ”.

Note: If they ask the same question with the same points with a symmetric about the x-axis then we have the equation of parabola is $$y^2=4ax$$ or $$y^2=-4ax$$ . Follow the same procedure as above you will get the equation of parabola. Careful about the points where it lies (quadrant). Remember the equations of parabola about x-axis and y-axis.