# Examples on Tangents and Normals To Rectangular Hyperbolas Set 1

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Example - 31

Show that, in general, four normals can be drawn to the hyperbola $$xy = {c^2}$$  from any point  $$P(h,\,k)$$ such that

(a) Sum of the x-coordinates of the feet of the normal = h

(b) Sum of the y-coordinates of the feet of the normal = k

(c) Product of the x-coordinates of the feet of the normal

= Product of the y-coordinates of the feet of the normal

=$$- {c^4}.$$

Solution: Any normal to  $$xy = {c^2}$$ can be written in the form

$x{t^3} - yt - c{t^4} + c = 0$

If this passes through  $$P(h,\,k),$$  we have

\begin{align}&\qquad\;\; h{t^3} - kt - c{t^4} + c = 0 \\ & \Rightarrow\quad c{t^4} - h{t^3} + kt - c = 0 \\\end{align}

This biquadratic equation gives (in general) four roots for t corresponding to the four normals. Let the four roots be  $${t_1},\,{t_2},\,{t_3},\,{t_4}.$$

We have,

 Sum of x-coordinates of feet of normals = \sum\limits_{i = 1}^4 {c{t_i} = c\sum\limits_{i = 1}^4 {{t_i} = c \cdot \begin{align}\frac{h}{c}\end{align} = h} } Sum of y-coordinates of feet of normals = \begin{align}\sum\limits_{i = 1}^4 {\frac{c}{{{t_i}}} = c\sum\limits_{i = 1}^4 {\frac{1}{{{t_i}}} = c \cdot \frac{k}{c} = k} } \end{align} Product of  x-coordinates of feet of normals = $${c^4}\prod\limits_{i = 1}^4 {{t_i} = - {c^4}}$$ Product of  y-coordinates of feet of normals = $${c^4}\prod\limits_{i = 1}^4 {\frac{1}{{{t_i}}} = - {c^4}}$$

This proves the assertions stated in the question.

Example - 32

A tangent is drawn to  $$xy = {c^2}$$  at the point P, intersecting the coordinates axes in A and B. Prove that area $$(\Delta OAB)$$  is constant where O is the origin.

Solution: Although we’ve already discussed the general problem(of which this problem is a particular example) in Example 29, we redo it here just to gain practice with tangents to $$xy = {c^2}.$$

Let the point P be  \left( {ct,\,\begin{align}\frac{c}{t}\end{align}} \right). The tangent at P has the equation

$x + y{t^2} = 2ct$

This intersects the x-axis in

$A \equiv (2ct,\,0)$

and the y - axis in

$B \equiv \left( {0,\,\frac{{2c}}{t}} \right)$

Thus, area (\Delta OAB) =\begin{align} \frac{1}{2} \times 2ct \times \frac{{2c}}{t}\end{align}

$$= 2{c^2}$$ (a constant)

Also notice that, as in Example -29, P is the mid-point of AB.

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