Question

Find the foci, vertices, directrices and axes of the following parabola $y = {x^2} - 2x + 3$.

Answer 1

Hint: When we come with these types of questions, the best way is to compare the given equation with the standard equation of that conic. From that you can get foci, directrix, latus rectum, axis and vertex easily.

Complete step by step answer:

Above equation can also be written as,

 $\Rightarrow y = {x^2} - 2x + 1 + 2$

 $\Rightarrow {(x - 1)^2} = (y - 2)$ - (Eq 1)

As we know, the standard equation of parabola is, 

${(x - {x_0})^2} = 4a(y - {y_0})$ where a is constant - (Eq 2) 

As we know that, foci of the standard parabola (equation 2) is, 

 $\Rightarrow$ foci (equation 2)= $({x_0},a + {y_0})$

 $\Rightarrow$ So, the foci of equation 1 will be,

On comparing equation 1 with equation 2,

 $\Rightarrow {x_0} = 1,{y_0} = 2$ and $a = \frac{1}{4}$

 $\Rightarrow$ So, foci (equation 1) = $(1,\frac{1}{4} + 2) = (1,2.25)$

As we know that vertex of the standard equation of parabola (equation 2) is,

 $\Rightarrow$ Vertex (equation 2) = $({x_0},{y_0})$

So, vertex of equation 1 will be, 

 $\Rightarrow$ Vertex (equation 1) = (1,2) 

Now, directrix of the standard parabola (equation 2) is $y = {y_0} - a$

So, directrix of the equation 1 will be, 

 $\Rightarrow$ Directrix (equation 1) is $y = 2 - \frac{1}{4} = 1.75$

And the axis of standard parabola (equation 2) is $x = {x_0}$

So, the axis of parabola (equation 1) is x = 1 

As you see equation 1 is plotted in the above graph.

 

Note: Understand the diagram properly whenever you are facing these kinds of problems. A better knowledge of formulas will be an added advantage.