Question

The equation $y^2-8y-x+19=0$ represents

• A. A parabola whose focus is $\left(\frac{1}{4},0\right)$ and directrix is $x=-\frac{1}{4}$
• B. A parabola whose vertex is $(3,4)$ and directrix is $x=\frac{11}{4}$
• C. A parabola whose focus is $\left(\frac{13}{4},4\right)$ and vertex is $(0,0)$
• D. A curve which is not a parabola

Answer 1

The correct option is B

A parabola whose vertex is $(3,4)$ and directrix is $x=\frac{11}{4}$

Explanation for the correct option:

Step 1: Standardize the given equation.

The equation $y^2-8y-x+19=0$ is given.

Standardize the given equation as follows:

$$\begin{gathered} y^2-8y-x+19=0 \\ \Rightarrow y^2-8y-x+16+3=0 \\ \Rightarrow {\left(y-4\right)}^2-x+3=0 \\ \Rightarrow {\left(y-4\right)}^2=x-3...1 \end{gathered}$$

Therefore, the Standardized equation is ${(y-4)}^2=x-3$. Which represents a parabola.

Step 2: Find the vertex and directrix of the parabola.

We know that the general equation of a parabola is ${(y-y_1)}^2=4a(x-x_1)...2$.

Where, $(x,y)$ is the general point of the parabola.

$(x_1,y_1)$ is the vertex of the parabola.

$4a$ is the length of the latus rectum.

$x=x_1-a$ is the equation of directrix.

On comparing equation $1$ and equation $2$, we get

The value of $a$ is $\frac{1}{4}$.

Vertex of the parabola is $(3,4)$

And the equation of directrix is $x=3-\frac{1}{4}$.

$\Rightarrow x=\frac{11}{4}$

Since the given equation represents a parabola whose vertex is at $(3,4)$ and the equation of directrix is $x=\frac{11}{4}$.

Hence, option $B$ is the correct .