Question

find the equation of parabola whose axis is parallel to y axis and which passes through the points (0,4) , (1,9) ,( -2,6). also find its latus rectum?

Answer 1

The general equation of a parabola is

$y=a{x}^2+bx+c$

where the axis is $x=-\frac{b}{2a}$

For point (0, 4):-

$4=a{\left(0\right)}^2+b\left(0\right)+c$

$\Rightarrow c=4$ -------------(1)

For point (1, 9):-

$9=a{\left(1\right)}^2+b\left(1\right)+c$

$$\begin{gathered} \Rightarrow a+b+4=9[U\sin g(1\left)\right] \\ \Rightarrow a+b=5--------\left(2\right) \end{gathered}$$

For point (-2, 6):-

$6=a{\left(-2\right)}^2+b\left(-2\right)+c$

$\Rightarrow 4a-2b+4=6[U\sin g(1\left)\right]$

$\Rightarrow 2a-b=1$ ----------------(3)

Eq. (2)+Eq. (3)$$\begin{gathered} \Rightarrow 3a=6 \\ \Rightarrow a=2 \end{gathered}$$

$\therefore Eq.\left(2\right)\Rightarrow b=3$

Hence, the parabola is $y=2x^2+3x+4$ and its axis is $x=-\frac{3}{4}$

Now, again we can write ​the form of the equation of the parabola with a horizontal axis is (x - h)² = 4p(y - k) where (h, k) is vertex and |4p| is the length of the latus rectum.

So, $y=2x^2+3x+4\leftrightarrow 4p(y-k)={\left(x-h\right)}^2$

$\Rightarrow 4\times \frac{1}{8}\left(y-\frac{23}{8}\right)={\left(x+\frac{3}{4}\right)}^2\leftrightarrow 4p(y-k)={\left(x-h\right)}^2$

$\Rightarrow p=\frac{1}{8}and\left(h,k\right)=\left(-\frac{3}{4},\frac{23}{8}\right)$

Now, the focus is at $\left(h,k+p\right)$

That is, at $\left(-\frac{3}{4},\frac{23}{8}+\frac{1}{8}\right),i.e.,\left(-\frac{3}{4},3\right)$

And the length of the latus rectum is 4p=$4\times \frac{1}{8}=\frac{1}{2}$