Question

The tangent at any point P on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with centre C,

meets the asymptotes in Q and R and cut off triangle CQR. Find the area of the triangle CQR.

• A. ab
• B. \)
• C. \)
• D.

Answer 1

The correct option is A

ab

Lets consider a parametric point P $(asec\theta ,btan\theta )$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1

Equation of the tangent at point P

$\frac{x.asec\theta }{a^2}-\frac{y.btan\theta }{b^2}=1$

$\frac{xsec\theta }{a}-\frac{ytan\theta }{b}=1$ -------------(1)

Equations of the asymptotes are

$y=\frac{b}{a}x$ -------------(2)

and $y=\frac{-b}{a}x$ --------------(3)

Let Q is the point of intersection of tangent of P with asymptotes $y=\frac{b}{a}x$

Solving equation (1) & (2)

$\frac{x.sec\theta }{a}-\frac{bxtan\theta }{a\times b}=1$

$\frac{x}{a}sec\theta -\frac{x}{a}tan\theta =1$

$\frac{x}{a}(sec\theta -tan\theta )=1$

$x=\frac{a}{(sin\theta -tan\theta )}$

$\Rightarrow$ $y=\frac{b}{a}\times \frac{a}{(sin\theta -tan\theta )}=\frac{b}{(sin\theta -tan\theta )}$

Coordinates of point P$(\frac{a}{(sin\theta -tan\theta )},\frac{b}{(sin\theta -tan\theta )})$

or

p$(\frac{a}{(sin\theta -tan\theta )}\times \frac{sec\theta +tan\theta }{(sin\theta +tan\theta )},\frac{b}{(sin\theta -tan\theta )}\times \frac{sec\theta +tan\theta }{(sin\theta +tan\theta )})$

$\Rightarrow p\left(a\right(sec\theta +tan\theta ),b(sec\theta +tan\theta )$

R is the point of intersection of tangent with h asymptotes $y=\frac{-b}{a}x$

similarly,

solving equation $\left(1\right)\&\left(3\right)$

we get $x=a(sec\theta -tan\theta )$

$y=-b(sec\theta -tan\theta )$

$R\left(a\right(sec\theta -tan\theta ),-b(sec\theta -tan\theta )$

Area of triangle CQR is

=$\frac{1}{2}$ $\left|\begin{array}{ccc}1& 1& 1\\ 0& a(sec\theta +tan\theta )& a(sec\theta -tan\theta )\\ 0& b(sec\theta +tan\theta )& b(sec\theta -tan\theta )\end{array}\right|$

$=\frac{1}{2}[1-ab(se{c}^2\theta -ta{n}^2\theta )-ab(se{c}^2-ta{n}^2\left)\right]$

$-1\times 0+1\times 0$

=$\frac{1}{2}[-2ab\times 1]$

=$|-ab|$

Area of the triangle CQR is ab