Name: Conjugate Hyperbola
Uploaded: 2019-07-19
Duration: 12 min 19 s
Description: In this video, we will learn about the transverse axis and conjugate axis of any hyperbola with respect to another hyperbola. Watch and learn.

Question

If

P_{1}, P_{2}

and

P_{3}

be three points on the rectangular hyperbola

x y = c^{2}

whose abscissae are

x_{1}, x_{2}

and

x_{3}

. Prove that the area of the triangle

P_{1} P_{2} P_{3}

is

\frac{c^{2}}{2} . |\frac{(x_{2} - x_{3}) (x_{3} - x_{1}) (x_{1} - x_{2})}{x_{1} x_{2} x_{3}}|

and that the tangents at these points forms a triangle whose area is,

2 c^{2} . |\frac{(x_{2} - x_{3}) (x_{3} - x_{1}) (x_{1} - x_{2})}{(x_{2} + x_{3}) (x_{3} + x_{1}) (x_{1} + x_{2})}|

Accepted Answer

Let $y_{1}, y_{2} & y_{3}$ are the ordinates of the points $P_{1}, P_{2} & P_{3}$ respectively.

Given $P_{1} (x_{1}, y_{1})$ , $P_{2} (x_{2}, y_{2})$ and $P_{3} (x_{3}, y_{3})$ be three points on the rectangular hyperbola $x y = c^{2} . . . (1)$ as shown in the figure.

Now converting $P_{1} (x_{1}, y_{1})$ into parametric form we get $P_{1} (c t_{1}, \frac{c}{t_{1}})$ .

Similarly, the coordinates of the other two points $P_{2} (c t_{2}, \frac{c}{t_{2}})$ and $P_{3} (c t_{3}, \frac{c}{t_{3}})$ .
The area of the triangle $P_{1} P_{2} P_{3}$ is given by

$a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} ||\begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix}||$

$\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} ||\begin{matrix} c t_{1} & \frac{c}{t_{1}} & 1 \\ c t_{2} & \frac{c}{t_{2}} & 1 \\ c t_{3} & \frac{c}{t_{3}} & 1 \end{matrix}||$

$\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} c^{2} ||\begin{matrix} t_{1} & \frac{1}{t_{1}} & 1 \\ t_{2} & \frac{1}{t_{2}} & 1 \\ t_{3} & \frac{1}{t_{3}} & 1 \end{matrix}||$ (taking $c$ common from the columns $C_{1} & C_{2}$ )

Applying row operations $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , we get

$\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} c^{2} ||\begin{matrix} t_{1} & \frac{1}{t_{1}} & 1 \\ t_{2} - t_{1} & \frac{1}{t_{2}} - \frac{1}{t_{1}} & 0 \\ t_{3} - t_{1} & \frac{1}{t_{3}} - \frac{1}{t_{1}} & 0 \end{matrix}||$

$\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} c^{2} ||\begin{matrix} t_{1} & \frac{1}{t_{1}} & 1 \\ t_{2} - t_{1} & \frac{t_{1} - t_{2}}{t_{1} t_{2}} & 0 \\ t_{3} - t_{1} & \frac{t_{1} - t_{3}}{t_{1} t_{3}} & 0 \end{matrix}||$

$\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} c^{2} |(t_{2} - t_{1}) (t_{3} - t_{1})| ||\begin{matrix} t_{1} & \frac{1}{t_{1}} & 1 \\ 1 & \frac{- 1}{t_{1} t_{2}} & 0 \\ 1 & \frac{- 1}{t_{1} t_{3}} & 0 \end{matrix}||$

Now expanding the determinant along the column $C_{3}$ , we get $\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} c^{2} |(t_{2} - t_{1}) (t_{3} - t_{1}) (\frac{- 1}{t_{1} t_{3}} + \frac{1}{t_{1} t_{2}})|$

$\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} c^{2} |(t_{2} - t_{1}) (t_{3} - t_{1}) (\frac{t_{3} - t_{2}}{t_{1} t_{2} t_{3}})|$

$\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} c^{2} |\frac{(t_{2} - t_{1}) (t_{3} - t_{1}) (t_{3} - t_{2})}{t_{1} t_{2} t_{3}}|$

$\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} c^{2} |\frac{(c t_{2} - c t_{1}) (c t_{3} - c t_{1}) (c t_{3} - c t_{2})}{(c t_{1}) (c t_{2}) (c t_{3})}| . . . (1)$ (Multiplying and dividing by $c^{3}$ )

Now using $c t_{1} = x_{1}, c t_{2} = x_{2}$ and $c t_{3} = x_{3}$ we get
$\Rightarrow a r (∆ P_{1} P_{2} P_{3}) = \frac{1}{2} c^{2} |\frac{(x_{2} - x_{1}) (x_{3} - x_{1}) (x_{3} - x_{2})}{x_{1} x_{2} x_{3}}|$

$a r (∆ P_{1} P_{2} P_{3}) = \frac{c^{2}}{2} |\frac{(x_{2} - x_{3}) (x_{3} - x_{1}) (x_{1} - x_{2})}{x_{1} x_{2} x_{3}}|$ . Hence, proved.

Again, the equation to the tangent at the point $P_{1} (c t_{1}, \frac{c}{t_{1}})$ with respect to the hyperbola will be $x + {t^{2}}_{1} y = 2 c t_{1} . . . (2)$

Similarly, tangents at $P_{2} (c t_{2}, \frac{c}{t_{2}})$ and $P_{3} (c t_{3}, \frac{c}{t_{3}})$ are $x + {t^{2}}_{2} y = 2 c t_{2} . . . (3)$ and $x + {t^{2}}_{3} y = 2 c t_{3} . . . (4)$

Solving the equations $(2)$ and $(3)$ , we get the point $A (\frac{2 c t_{1} t_{2}}{t_{1} + t_{2}}, \frac{2 c}{t_{1} + t_{2}})$ as shown in the figure. Similarly, we can get the coordinates of $B (\frac{2 c t_{2} t_{3}}{t_{2} + t_{3}}, \frac{2 c}{t_{2} + t_{3}})$ and $C (\frac{2 c t_{1} t_{3}}{t_{1} + t_{3}}, \frac{2 c}{t_{1} + t_{3}})$
Hence, the area of the triangle $A B C$ is given by $a r (∆ A B C) = \frac{1}{2} ||\begin{matrix} \frac{2 c t_{1} t_{2}}{t_{1} + t_{2}} & \frac{2 c}{t_{1} + t_{2}} & 1 \\ \frac{2 c t_{2} t_{3}}{t_{2} + t_{3}} & \frac{2 c}{t_{2} + t_{3}} & 1 \\ \frac{2 c t_{1} t_{3}}{t_{1} + t_{3}} & \frac{2 c}{t_{1} + t_{3}} & 1 \end{matrix}||$

Applying $R_{1} \to (t_{1} + t_{2}) R_{1}, R_{2} \to (t_{2} + t_{3}) R_{2}, R_{3} \to (t_{1} + t_{3}) R_{3}$ , we get

$a r (∆ A B C) = \frac{1}{2} \frac{1}{(t_{1} + t_{2}) (t_{1} + t_{2}) (t_{1} + t_{3})} ||\begin{matrix} 2 c t_{1} t_{2} & 2 c & t_{1} + t_{2} \\ 2 c t_{2} t_{3} & 2 c & t_{2} + t_{3} \\ 2 c t_{1} t_{3} & 2 c & t_{1} + t_{3} \end{matrix}||$

$\Rightarrow a r (∆ A B C) = \frac{1}{2} \frac{4 c^{2}}{(t_{1} + t_{2}) (t_{1} + t_{2}) (t_{1} + t_{3})} ||\begin{matrix} t_{1} t_{2} & 1 & t_{1} + t_{2} \\ t_{2} t_{3} & 1 & t_{2} + t_{3} \\ t_{1} t_{3} & 1 & t_{1} + t_{3} \end{matrix}||$ (taking common $2 c$ from the columns $C_{1} & C_{2}$ )

Applying row operations $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , we get

$a r (∆ A B C) = \frac{1}{2} |\frac{4 c^{2}}{(t_{1} + t_{2}) (t_{1} + t_{2}) (t_{1} + t_{3})}| ||\begin{matrix} t_{1} t_{2} & 1 & t_{1} + t_{2} \\ t_{2} (t_{3} - t_{1}) & 0 & t_{3} - t_{1} \\ t_{1} (t_{3} - t_{2}) & 0 & t_{3} - t_{2} \end{matrix}||$

$\Rightarrow a r (∆ A B C) = \frac{1}{2} |\frac{4 c^{2} (t_{3} - t_{1}) (t_{3} - t_{2})}{(t_{1} + t_{2}) (t_{1} + t_{2}) (t_{1} + t_{3})}| ||\begin{matrix} t_{1} t_{2} & 1 & t_{1} + t_{2} \\ t_{2} & 0 & 1 \\ t_{1} & 0 & 1 \end{matrix}||$ (taking common $(t_{3} - t_{1}), (t_{3} - t_{2})$ from the rows $R_{2} & R_{3}$ respectively)

Now expanding the determinant along the column $C_{2}$ , we get

$a r (∆ A B C) = 2 c^{2} |\frac{(c t_{3} - c t_{1}) (c t_{3} - c t_{2}) (c t_{2} - c t_{1})}{(c t_{1} + c t_{2}) (c t_{1} + c t_{2}) (c t_{1} + c t_{3})}| . . . (5)$

(Multiplying and dividing by $c^{3}$ )

Now using $c t_{1} = x_{1}, c t_{2} = x_{2}$ and $c t_{3} = x_{3}$ in $(5)$ , we get $a r (∆ A B C) = 2 c^{2} |\frac{(x_{1} - x_{2}) (x_{2} - x_{3}) (x_{3} - x_{1})}{(x_{2} + x_{3}) (x_{3} + x_{1}) (x_{1} + . x_{2})}| .$ Hence, proved.

Important Questions on The Hyperbola

Learn and Prepare for any exam you want !

If P1,P2 and P3 be three points on the rectangular hyperbola xy=c2 whose abscissae are x1,x2 and x3. Prove that the area of the triangle P1P2P3 is c22.x2-x3x3-x1x1-x2x1x2x3 and that the tangents at these points forms a triangle whose area is,2c2.x2-x3x3-x1x1-x2x2+x3x3+x1x1+x2

Important Questions on The Hyperbola

Learn and Prepare for any exam you want !