- #1

issacnewton

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## 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