Given the equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a \neq b$, we need to find the equation of the locus of points from which two tangents can be drawn to the ellipse such that: (i) the slopes of the two tangents are reciprocals of each other, and (ii) the slopes of the two tangents are negative reciprocals of each other.

1. Problem Description

Given the equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a \neq b$, we need to find the equation of the locus of points from which two tangents can be drawn to the ellipse such that:

(i) the slopes of the two tangents are reciprocals of each other, and

(ii) the slopes of the two tangents are negative reciprocals of each other.

2. Solution Steps

Let $(h, k)$ be a point from which tangents are drawn to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The equation of a tangent with slope $m$ is given by

$y = mx \pm \sqrt{a^2m^2 + b^2}$.

If this tangent passes through $(h, k)$, then

$k = mh \pm \sqrt{a^2m^2 + b^2}$.

Rearranging the terms and squaring, we have

$(k - mh)^2 = a^2m^2 + b^2$

$k^2 - 2mhk + m^2h^2 = a^2m^2 + b^2$

$(h^2 - a^2)m^2 - 2hkm + (k^2 - b^2) = 0$.

This is a quadratic equation in $m$. Let $m_1$ and $m_2$ be the roots of this equation, which represent the slopes of the two tangents.

(i) If $m_1$ and $m_2$ are reciprocals, then $m_1m_2 = 1$.

From the quadratic equation, the product of the roots is given by

$m_1m_2 = \frac{k^2 - b^2}{h^2 - a^2}$.

Since $m_1m_2 = 1$, we have

$\frac{k^2 - b^2}{h^2 - a^2} = 1$

$k^2 - b^2 = h^2 - a^2$

$h^2 - k^2 = a^2 - b^2$.

Replacing $(h, k)$ with $(x, y)$, we get the equation of the locus as $x^2 - y^2 = a^2 - b^2$.

(ii) If $m_1$ and $m_2$ are negative reciprocals, then $m_1m_2 = -1$.

$\frac{k^2 - b^2}{h^2 - a^2} = -1$

$k^2 - b^2 = -h^2 + a^2$

$h^2 + k^2 = a^2 + b^2$.

Replacing $(h, k)$ with $(x, y)$, we get the equation of the locus as $x^2 + y^2 = a^2 + b^2$.

3. Final Answer

(i) If the slopes are reciprocals, the equation is $x^2 - y^2 = a^2 - b^2$.

(ii) If the slopes are negative reciprocals, the equation is $x^2 + y^2 = a^2 + b^2$.