# Examples on Tangents and Normals To Rectangular Hyperbolas Set 1

**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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