Find the equation of parabola with vertex (2, 1) and focus (1, -1)

[sagarsoni7]

Hello friends. This is my first post. I am given the following problem:

Find the equation of parabola with vertex (2, 1) and focus (1, -1)

I have tried to solve this question in this way:

Since vertex is mid point of Focus and the point which touches the Directrix. From this, I calculate that point which touches the directrix and it came out to be (3, 3). From this I got equation of directrix to be y-3=0. Now let any point (x, y) on parabola. Now distance from focus to this point and the perpendicular distance of this point from directrix will be equal. From this I got the equation: x^2 - 2x -4y + 11 =0. But answer given is: 4 x^2 + y^2 - 4xy +8x+ 46y -71=0. Please tell me where is my mistake? If you have any other method to solve this question, please share. Thanks in advance.

Sagar Soni
Class 11th Student
India

 

[MarkFL]

Hello and welcome to MHB, sagarsoni7! (Wave)

I have added the problem statement to the body of your post so that it is more clear to those reading. :D

Now, I think your method is good, but you have made an error in calculating the equation of the directrix.

It will pass through the point $(3,3)$ and be perpendicular to the axis of symmetry, which will pass through the focus and the vertex. Can you find the slope of the directrix from this? And then use this slope and the point we found to determine the equation of the directrix?

 

[sagarsoni7]

Hello and welcome to MHB, sagarsoni7! (Wave)

I have added the problem statement to the body of your post so that it is more clear to those reading. :D

Now, I think your method is good, but you have made an error in calculating the equation of the directrix.

It will pass through the point $(3,3)$ and be perpendicular to the axis of symmetry, which will pass through the focus and the vertex. Can you find the slope of the directrix from this? And then use this slope and the point we found to determine the equation of the directrix?

Thanks mate. You were right. Now I have solved the problem.

 

[MarkFL]

For the benefit of others who might search for the solution to this type of problem, I thought I would post a method for getting the solution.

First, we find the slope $m$ of the axis of symmetry:

\(\displaystyle m=\frac{1-(-1)}{2-1}=2\)

Thus, the directrix, being perpendicular to the axis of symmetry will have a slope of \(\displaystyle -\frac{1}{2}\)

We know the directrix passes through the point $(3,3)$, and so using the point-slope formula, we can find the equation of the directrix as follows:

\(\displaystyle y-3=-\frac{1}{2}(x-3)\)

Arranging this in slope-intercept form, we obtain:

\(\displaystyle y=-\frac{1}{2}x+\frac{9}{2}\)

Now, the parabola we are seeking will, by definition, be the locus of points equidistant from the directrix and the focus. And so using the formulas for the distance between two points and between a point and a line, we may write:

\(\displaystyle (x-1)^2+(y+1)^2=\frac{\left(-\frac{1}{2}x+\frac{9}{2}-y\right)^2}{\left(-\frac{1}{2}\right)^2+1}\)

\(\displaystyle (x-1)^2+(y+1)^2=\frac{\left(9-x-2y\right)^2}{5}\)

\(\displaystyle 5x^2-10x+5+5y^2+10y+5=x^2+4xy-18x+4y^2-36y+81\)

\(\displaystyle 4x^2+y^2-4xy+8x+46y-71=0\)

 

[CellCoree]

Hello Sagar,

Great job on attempting to solve this problem! The equation you have found is actually correct, but it is not in the standard form of a parabola. The standard form for a parabola with vertex (h, k) and focus (p, q) is (x-h)^2=4p(y-k).

So in this case, the equation would be (x-2)^2=4(1)(y-1), which simplifies to x^2-4x+4=4y-4. Rearranging this equation, we get x^2-4x-4y+8=0, which is equivalent to the answer given to you.

Keep up the great work in your studies and keep practicing! Best of luck to you.

Sincerely,