Find locus of points near ellipse with some condition

[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

 

[haruspex]

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.

 

[issacnewton]

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

 

[haruspex]

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.

 

[issacnewton]

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

 

[haruspex]

which is a hyperbola

in particular, a rectangular hyperbola