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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