Equation of parabola when equation of axis and tangent at vertex and latusrectum are given - Practice Questions & MCQ

2nd form of Parabola and Shifted Parabola

2nd form of Parabola

Equation of parabola when the equation of axis, tangent at the vertex and latus rectum length are given
Let the equation of the axis be $\mathrm{lx}+\mathrm{my}+\mathrm{n}=0$ and the equation of the tangent at the vertex be $\mathrm{mx}-\mathrm{ly}+\mathrm{r}=0$.

\begin{aligned}
&\text { The equation of parabola is }\\
&\begin{gathered}
(\mathrm{PM})^2=(\text { Latusrectum }) \cdot(\mathrm{PN}) \\
\Rightarrow\left(\frac{\mathrm{lx}+\mathrm{my}+\mathrm{n}}{\sqrt{\mathrm{l}^2+\mathrm{m}^2}}\right)^2=(\text { Latusrectum }) \cdot\left(\frac{\mathrm{mx}-\mathrm{ly}+\mathrm{r}}{\sqrt{\mathrm{~m}^2+\mathrm{l}^2}}\right)
\end{gathered}
\end{aligned}

Shifted Parabola

If the parabola $\mathbf{y}^{\mathbf{2}} \mathbf{= 4 an x}$ is shifted (without any rotation) to a new position with a new vertex as $(p, q)$, then the equation of a new parabola is $(\mathbf{y}-q)^{\mathbf{2}}=\mathbf{4 a}(\mathbf{x}-\mathrm{p})$
If the parabola $\mathbf{y}^{\mathbf{2}}=\mathbf{4 an x}$ is shifted (without any rotation) to a new position with a new vertex as $(p, q)$, then the equation of a new parabola is $(\mathbf{y}-q)^{\mathbf{2}}=\mathbf{4 a ( x p )}$
If the parabola $x^2=4 a y$ is shifted (without any rotation) to a new position with a new vertex as $(p, q)$, then the equation of a new parabola is $(x-p)^2=4 a(y-q)$
If the parabola $x^2=-4 a y$ is shifted (without any rotation) to a new position with a new vertex as $(p, q)$, then the equation of a new parabola is $(x-p)^2=-4 a(y-q)$

Note: Parametric point of $(y-q)^2=4 a(x-p)$ is $\left(p+a t^2, q+2 a t\right)$

Equation of a Parabola When The Vertex is (h k ) and Parabolic Curve 

The equation of the parabola when the axis is parallel to the x-axis 

$
\mathrm{y}^2=4 \mathrm{ax}
$

can be written as $(\mathrm{y}-0)^2=4 \mathrm{a}(\mathrm{x}-0)$
The vertex of the parabola is $O(0,0)$. Now the origin is shifted to $\mathrm{O}^{\prime}(h, k)$ without changing the direction of axes, its equation becomes $(y-k)^2=4 a(x-h)$

The parametric equation of the curve $(y-k)^2=4 a(x-h)$ are $x-h+a t^2$ and $y=k+2 a t$

Thus its focus is $S(a+h, k)$, latus rectum $=4 a$ and the equation of the directrix is $x=h-a$, i.e. $x+a-h=0$

Another equation of the parabola when the axis is parallel to the $y$-axis is

$
(x-h)^2=4 a(y-k)
$

Its focus is $S(h, a+k)$, latus rectum $=4 a$ and the equation of the directrix is $y=k-a$, i.e. $y+a-k=0$

Parabolic Curve
The equations of the form $\mathrm{y}=\mathrm{Ax}^2+\mathrm{Bx}+\mathrm{C}$ and $\mathrm{x}=\mathrm{Ay}^2+\mathrm{By}+\mathrm{C}$ are always represents parabolas, generally called as parabolic curve.
Now,

$
\begin{aligned}
y & =A x^2+B x+C \\
& =A\left(x^2+\frac{B}{A} x+\frac{C}{A}\right) \\
& =A\left\{\left(x+\frac{B}{2 A}\right)^2-\frac{B^2}{4 A^2}+\frac{C}{A}\right\} \\
& =A\left\{\left(x+\frac{B}{2 A}\right)^2-\frac{\left(B^2-4 A C\right)}{4 A^2}\right\}
\end{aligned}
$
Above equation can be written as

$
\left(x+\frac{B}{2 A}\right)^2=\frac{1}{A}\left(y+\frac{\left(B^2-4 A C\right)}{4 A^2}\right)
$

comparing this equation with $(\mathrm{x}-\mathrm{h})^2=4 \mathrm{a}(\mathrm{y}-\mathrm{k})$ it represent a parabola
with the vertex $(\mathrm{h}, \mathrm{k})=\left(-\frac{\mathrm{B}}{2 \mathrm{~A}},-\frac{\mathrm{B}^2-4 \mathrm{AC}}{4 \mathrm{~A}}\right)$ and axis parallel to Y - axis
and latusrectum $=\frac{1}{|\mathrm{~A}|}$.