Question

Write an equation of the horizontal parabola with the given vertex $\left( -3,-5 \right)$ and passing through the given point $\left( -5,-6 \right)$?

Answer 1

Hint: We start solving the problem by recalling the equation of the horizontal parabola is ${{\left( y-q \right)}^{2}}=4a\left( x-p \right)$, where the $\left( p,q \right)$ is the vertex of the parabola and ‘4a’ is the length of the latus rectum. We then compare the given vertex with $\left( p,q \right)$ to find the values of ‘p’ and ‘q’. We then substitute the given point $\left( -5,-6 \right)$ to find the value of ‘a’, which will give us the required equation of the parabola.

Complete step by step answer:
According to the problem, we need to find the equation of the horizontal parabola with the vertex $\left( -3,-5 \right)$ and passing through the point $\left( -5,-6 \right)$.

We know that the equation of the horizontal parabola is ${{\left( y-q \right)}^{2}}=4a\left( x-p \right)$, where the $\left( p,q \right)$ is the vertex of the parabola and ‘4a’ is the length of the latus rectum.
Let us compare $\left( -3,-5 \right)$ with $\left( p,q \right)$. So, we get $p=-3$ and $q=-5$.
So, the equation of the parabola is ${{\left( y+5 \right)}^{2}}=4a\left( x+3 \right)$. We know that this parabola is passing through the point $\left( -5,-6 \right)$.
Let us substitute this point in ${{\left( y+5 \right)}^{2}}=4a\left( x+3 \right)$ to get the required equation.
So, we get ${{\left( -6+5 \right)}^{2}}=4a\left( -5+3 \right)$.
$\Rightarrow {{\left( -1 \right)}^{2}}=4a\left( -2 \right)$.
$\Rightarrow 1=-8a$.
$\Rightarrow a=\dfrac{-1}{8}$. Let us substitute this value in the equation ${{\left( y+5 \right)}^{2}}=4a\left( x+3 \right)$.
So, the equation of the parabola is ${{\left( y+5 \right)}^{2}}=4\left( \dfrac{-1}{8} \right)\left( x+3 \right)$.
$\Rightarrow {{\left( y+5 \right)}^{2}}=\left( \dfrac{-1}{2} \right)\times \left( x+3 \right)$.

So, we have found the equation of the parabola as ${{\left( y+5 \right)}^{2}}=\left( \dfrac{-1}{2} \right)\times \left( x+3 \right)$.

Note: We should know that the horizontal parabola has an axis parallel to the x-axis and the directrix is parallel to the y-axis. Whereas the vertical parabola has an axis parallel to the y-axis and the directrix parallel to the x-axis. We should know that the equation of the vertical parabola is similar to the quadratic equation of degree 2. Similarly, we can expect to find the equation of the tangent and normal at the vertex of the obtained parabola.