Question

Find vertex, axis, focus, directrix, latus rectum of the parabola $4{y^2} + 12x - 20y + 67 = 0$.
A. Vertex: $\left( { - \dfrac{7}{2},\dfrac{5}{2}} \right)$, Axis: $y = \dfrac{5}{2}$, Focus: $\left( { - \dfrac{{17}}{4},\dfrac{5}{2}} \right)$, Directrix: $x = - \dfrac{{11}}{4}$, LR: $ = 3$
B. Vertex: $\left( { - \dfrac{9}{2},\dfrac{5}{2}} \right)$, Axis: $y = \dfrac{5}{2}$, Focus: $\left( { - \dfrac{{17}}{4},\dfrac{5}{2}} \right)$, Directrix: $x = - \dfrac{{11}}{4}$, LR: $ = 4$
C. Vertex: $\left( { - \dfrac{9}{2},\dfrac{5}{2}} \right)$, Axis: $y = \dfrac{5}{2}$, Focus: $\left( { - \dfrac{{17}}{4},\dfrac{5}{2}} \right)$, Directrix: $x = - \dfrac{{11}}{4}$, LR: $ = 3$
D. Vertex: $\left( { - \dfrac{7}{2},\dfrac{5}{2}} \right)$, Axis: $y = \dfrac{5}{2}$, Focus: $\left( { - \dfrac{{17}}{4},\dfrac{5}{2}} \right)$, Directrix: $x = - \dfrac{{11}}{4}$, LR: $ = 4$

Answer 1

Hint:We are provided with an equation of parabola in the given problem. Then, we should compare the equation with the general equation of parabola in order to determine the vertex and focus of the given equation, and also to determine their orientation. Then we will use the formula of directrix to find the directrix of the conic section represented by the given equation. We should also know that if the origin is shifted to $\left( {h,k} \right)$, then X becomes $\left( {x - h} \right)$ and Y becomes $\left( {y - k} \right)$.

Complete step by step answer:
Given the equation: $4{y^2} + 12x - 20y + 67 = 0$
Then, we group the terms consisting of y. So, we have,
$ \Rightarrow \left( {4{y^2} - 20y} \right) + 12x + 67 = 0$
We factor out $4$ from the equation. So, we have,
$ \Rightarrow 4\left( {{y^2} - 5y} \right) + 12x + 67 = 0$
Now, we know the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$. So, we get,
$ \Rightarrow 4\left( {{{\left( {y - \dfrac{5}{2}} \right)}^2} - \dfrac{{25}}{4}} \right) + 12x + 67 = 0$
Now, opening the bracket and simplifying the equation, we get,
$ \Rightarrow 4{\left( {y - \dfrac{5}{2}} \right)^2} - 25 + 12x + 67 = 0$
$ \Rightarrow 4{\left( {y - \dfrac{5}{2}} \right)^2} + 12x + 42 = 0$
Now, we take all the terms not containing y to the right side of equation,
$ \Rightarrow 4{\left( {y - \dfrac{5}{2}} \right)^2} = - 12x - 42$
Dividing both sides of equation by four, we get,
$ \Rightarrow {\left( {y - \dfrac{5}{2}} \right)^2} = - \dfrac{{\left( {12x + 42} \right)}}{4}$
$ \Rightarrow {\left( {y - \dfrac{5}{2}} \right)^2} = - 3\left( {x + \dfrac{7}{2}} \right) - - - - \left( 1 \right)$

The degree of the term containing y is $2$ and x is $1$. So, the parabola must be in horizontal orientation. Since the coefficient of the term consisting of x is negative, so the parabola must be facing leftwards. General form of a horizontal parabola facing leftwards is
${\left( {y - k} \right)^2} = \left( { - 4a} \right)\left( {x - h} \right)$
Comparing equation $(1)$ with the general form of equation, we get,
$h = - \dfrac{7}{2}$
$\Rightarrow k = \dfrac{5}{2}$
$\Rightarrow - 4a = - 3$
So, the coordinates of the vertex of the parabola are: $\left( {h,k} \right)$.
So, the coordinates of the vertex of the given parabola are: $\left( { - \dfrac{7}{2},\dfrac{5}{2}} \right)$.

Now, we also have $ - 4a = - 3$.
Dividing both sides of the equation by $ - 4$.
$ \Rightarrow a = \dfrac{3}{4}$
So, the value of a is $\left( {\dfrac{3}{4}} \right)$.
Now, for a parabola facing leftwards having vertex at origin, the directrix has an equation $x = a$. But the parabola given to us has a vertex at $\left( { - 2,2} \right)$. So, we use the concept of shifting of origin to find the directrix of this parabola. So, we have, equation of directrix as: $X = a$

Now, we replace X with $\left( {x - h} \right)$ to find the equation of the directrix for this shifted parabola. So, we get,
$ \Rightarrow x - h = a$
Putting in the value of a and h, we get,
$ \Rightarrow x - \left( { - \dfrac{7}{2}} \right) = \dfrac{3}{4}$
Simplifying the equation,
$ \Rightarrow x + \dfrac{7}{2} = \dfrac{3}{4}$
\[ \Rightarrow x = \dfrac{3}{4} - \dfrac{7}{2}\]
Taking LCM of denominators,
\[ \Rightarrow x = \dfrac{3}{4} - \dfrac{{14}}{4}\]
\[ \Rightarrow x = - \dfrac{{11}}{4}\]
Hence, the equation of the directrix of the parabola is \[x = - \dfrac{{11}}{4}\]. Similarly, the coordinates of focus of a leftwards facing parabola are $\left( { - a,0} \right)$. But the parabola given to us has a vertex at $\left( { - 2,2} \right)$. So, we use the concept of shifting of origin to find the coordinates of focus. So, we have, coordinates of focus as: $X = - a$ and $Y = 0$

Now, we replace the X with $\left( {x - h} \right)$ and Y with $\left( {y - k} \right)$ to find the coordinate focus of the parabola.
So, we get,
$ \Rightarrow x - h = - a$ and $y - k = 0$
Putting in the value of a, h and k we get,
$ \Rightarrow x - \left( { - \dfrac{7}{2}} \right) = - \dfrac{3}{4}$ and $y - \dfrac{5}{2} = 0$
$ \Rightarrow x + \dfrac{7}{2} = - \dfrac{3}{4}$ and $y = \dfrac{5}{2}$
$ \Rightarrow x = - \dfrac{{17}}{4}$ and $y = \dfrac{5}{2}$
So, the coordinate focus of the parabola are: $\left( { - \dfrac{{17}}{4},\dfrac{5}{2}} \right)$.
Now, length of latus rectum is equal to $\left| {4a} \right|$ units. So, we know that the value of $a$ is $ - \dfrac{3}{4}$.So, we get the latus rectum as $\left| {4\left( { - \dfrac{3}{4}} \right)} \right| = \left| { - 3} \right| = 3$ units.

Now, we know that the axis of a horizontal parabola having vertex at origin is $y = 0$.But the parabola given to us has a vertex at $\left( { - 2,2} \right)$. So, we use the concept of shifting of origin to find the equation of the axis of parabola. So, we have, equation of the axis of the parabola as $Y = 0$. Now, we replace Y with $\left( {y - k} \right)$ to find the coordinates of the focus of the parabola.
So, $y - k = 0$
Putting the value of k in the expression, we get,
 $ \Rightarrow y = k = \dfrac{5}{2}$
So, the axis of the parabola is \[y = \dfrac{5}{2}\].

Hence, option A is the correct answer.

Note: For a parabola, the variable with degree $2$ gives us the idea, whether the parabola is symmetric to x-axis or y-axis. Then, depending upon the sign of a, we can determine the orientation of the parabola. Like, for a parabola symmetric to x-axis, if a is positive, then the parabola is right handed, and if a is negative, then the parabola is left handed. And, for a parabola symmetric to y-axis, if a is positive, then the parabola is open upwards, and if a is negative, then the parabola is open downwards. In this way, we can determine the orientation of the parabola. We must know the concept of shifting of origin in order to solve the given question.