Browse Questions

Prove that the tangent at any point to the rectangular hyperbola forms with the asymptotes a tringle of constant area.

Toolbox:
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• For the standard rectangular hyperbola,the coordinate axes are taken to the asymptotes and the equation of the hyperbola is $xy=c^2$ .
• If the centre is at $(h,k)$ with asymptotes parallel to the coordinate axes,the equation is $(x-h)(y-k)=c^2$
Step 1:
Let the equation of the rectangular hyperbola be $xy=c^2$ and let $P(x_1,y_1)$ be any point on it.
Therefore $x_1y_1=c^2$
The tangent at $P$ is $\large\frac{1}{2}$$(xy_1+yx_1)=c^2 (i.e) xy_1+yx_1=2c^2 Step 3: It intersects the x and y-axes at A and B respectively. The x-coordinates of A and y-coordinates of B are obtained by y=0 and x=0 in the equation of the tangent.A and B are the points A(\large\frac{2c^2}{y_1},$$0),B(0,\large\frac{2c^2}{x_1})$
Step 4:
Now $OA=\large\frac{2c^2}{y_1},$$OB=\large\frac{2c^2}{x_1} and \Delta AOB is rightangled at O. Its area=\large\frac{1}{2}$$OA\times OB$
$\qquad\;\;=\large\frac{1}{2}$$\times\large\frac{2c^2}{y_1}\times$$\large\frac{2c^2}{x_1}$
$\Rightarrow \large\frac{2c^4}{x_1y_1}$
We know that $x_1y_1=c^2$
$\Rightarrow \large\frac{2c^4}{c^2}$
$\Rightarrow 2c^2$
Therefore the area of the $\Delta$ formed is a constant.