# Find locus of points near ellipse with some condition

• issacnewton
In summary, the given problem involves finding the equation of the set of all points from which there are two tangents to an ellipse with specific slopes. After solving the equations and eliminating variables, it is found that for the slopes to be reciprocals, the locus of points is a hyperbola, while for the slopes to be negative reciprocals, the locus of points is a circle.
issacnewton

## Homework Statement

Given an ellipse ##\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1## , where ##a \ne b##, find the equation of the set of all points from which there are two tangents to the curve whose slopes are (a) reciprocals and (b) negative reciprocals

## Homework Equations

Equation of a tangent to an ellipse at a point ##(x_o, y_o)## is ##\frac{x x_o}{a^2} + \frac{y y_o}{b^2}=1##

## The Attempt at a Solution

Here is my attempt for the solution for (a). Let ##(h,k)## be the point from where we will have the two tangents to the ellipse ##\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1##. Let these two tangents meet the ellipse at points ##(x_1,y_1)## and ##(x_2, y_2)##. So the equations of these tangents are ##\frac{x x_1}{a^2} + \frac{y y_1}{b^2}=1## and ##\frac{x x_2}{a^2} + \frac{y y_2}{b^2}=1##. Now ##(h,k)## lies on these tangents. So we must have ##\frac{h x_1}{a^2} + \frac{k y_1}{b^2}=1## and ##\frac{h x_2}{a^2} + \frac{k y_2}{b^2}=1##. Also the slopes of these tangents are ##m_1 = -\left( \frac{b^2 x_1}{a^2 y_1} \right)## and ##m_2 = -\left( \frac{b^2 x_2}{a^2 y_2} \right)##. Since the product of the slopes is 1, we must have $$\left( \frac{b^2 x_1}{a^2 y_1} \right)\left( \frac{b^2 x_2}{a^2 y_2} \right) = 1\cdots\cdots(1)$$ Also the points ##(x_1, y_1)## and ##(x_2, y_2)## lie on the ellipse, so they satisfy the equation of the ellipse $$\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1\cdots\cdots(2)$$
$$\frac{x_2^2}{a^2} + \frac{y_2^2}{b^2} = 1\cdots\cdots(3)$$ After this I did lot of algebra and tried to get an equation involving only ##h,k,a,b## , but I am unable to eliminate the variables ##x_1,y_1,x_2,y_2##. I think at this point I have all the information I need to solve this problem. Any hints to further this problem will be helpful.
Thanks

IssacNewton said:
I am unable to eliminate the variables ##x_1,y_1,x_2,y_2##.
You have five equations, so you should in principle be able to eliminate those vars and have the one equation left.
I tried substituting y1=b sin(θ1) etc. It made the algebra a lot easier, but I just ended up with the problem of eliminating the two angles.

Ok, Haruspex I will try to use angles and see if I can get any improvement.

IssacNewton said:
Ok, Haruspex I will try to use angles and see if I can get any improvement.
For what it's worth, my guess is rectangular hyperbola.

Ok, I found the solution and I am presenting here. Ellipse is ##\frac{x^2}{a^2}+\frac{y^2}{b^2}=1##. Let the tangent to ellipse be ##y = mx+c##. At the point of intersection, both of these equations must be satisfied. So we have ##\frac{x^2}{a^2} + \frac{(mx+c)^2}{b^2}=1##. This can be simplified as $$(b^2 + m^2a^2)x^2 + (2mca^2)x + (a^2c^2 - a^2b^2) = 0\cdots\cdots(1)$$ Now tangent touches at a single point to the ellipse, so roots of the above quadratic equation can not be distinct. Fo this to happen, we must have discriminant equal to ##0##. So we have $$(2mca^2)^2 - 4(b^2 + m^2a^2)(a^2c^2- a^2b^2)=0 \cdots\cdots(2)$$ Simplifying this, we have ##c^2 = b^2 + m^2a^2##. Now let ##(h,k)## be an external point to the ellipse from where we draw the tangent. So ##(h,k)## must lie on the line ##y = mx+c##. Therefore ##k = mh+c##. This leads to $$(k-mh)^2 = c^2 = b^2+m^2a^2$$ Expanding, we get $$(h^2- a^2)m^2 - 2mhk + (k^2 - b^2) = 0$$ This is a quadratic equation in ##m##, since we have two tangents from an external point to this ellipse. Now from the theory of quadratic equations, the product of roots is given by $$m_1m_2 = \frac{k^2 - b^2}{h^2 - a^2}$$ Now in part (a), this product is equal to 1. This leads us to ##h^2 - k^2 = a^2 - b^2##. So the locus of points here is ##x^2 - y^2 = a^2 - b^2##, which is a hyperbola. For part (b), the product is equal to -1, which leads to ##h^2 + k^2 = a^2 + b^2##. So the locus of points here is ##x^2 + y^2 = a^2 + b^2##. This is a circle. I found a question on stack exchange which helped me solve this problem. Here is the Link

Thanks

IssacNewton said:
which is a hyperbola
in particular, a rectangular hyperbola

## 1. What is the definition of a locus of points near an ellipse?

The locus of points near an ellipse refers to the set of all points that satisfy a given condition and are located within a certain distance of the ellipse.

## 2. How is the locus of points near an ellipse calculated?

The locus of points near an ellipse can be calculated using mathematical equations and principles, such as the Pythagorean theorem and the distance formula. It also requires knowledge of the ellipse's center, major and minor axes, and the given condition.

## 3. What are some common conditions used to find the locus of points near an ellipse?

Some common conditions used in finding the locus of points near an ellipse include being a certain distance from the ellipse, being tangent to the ellipse, or intersecting the ellipse at a right angle.

## 4. How does the shape and orientation of the ellipse affect the locus of points near it?

The shape and orientation of the ellipse can greatly affect the locus of points near it. For example, a more elongated ellipse will have a larger locus of points compared to a more circular ellipse. Similarly, the orientation of the ellipse can also impact the shape and size of the locus of points.

## 5. What are some real-world applications of finding the locus of points near an ellipse?

Finding the locus of points near an ellipse has many practical applications, such as in astronomy for determining the possible orbits of celestial bodies, in engineering for designing curved structures, and in navigation for determining the possible locations of objects based on their distance from a known point.

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