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

AI Thread Summary
The discussion revolves around finding the equation of a parabola with a given vertex at (2, 1) and focus at (1, -1). The initial attempt involved calculating the directrix incorrectly, leading to an erroneous equation. After guidance, the correct slope of the directrix was determined to be -1/2, allowing for the accurate formulation of its equation. The final equation of the parabola was derived as 4x^2 + y^2 - 4xy + 8x + 46y - 71 = 0. This solution highlights the importance of correctly identifying the directrix in parabola problems.
sagarsoni7
Messages
2
Reaction score
0
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
 
Mathematics news on Phys.org
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?
 
MarkFL said:
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.
 
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:

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

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

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:

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

Arranging this in slope-intercept form, we obtain:

$$y=-\frac{1}{2}x+\frac{9}{2}$$

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:

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

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

$$5x^2-10x+5+5y^2+10y+5=x^2+4xy-18x+4y^2-36y+81$$

$$4x^2+y^2-4xy+8x+46y-71=0$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top