Question

A tangent to a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ intercepts a length of unity from each of the coordinate axes, then the point (a, b) lies on the rectangular hyperbola
(A)$ {x^2}-{y^2}=2$
(B)${ x^2}-{y^2}=1$
(C)$ {x^2}-{y^2}=-1$
(D) None of these

Answer 1

Hint: In analytic geometry, a hyperbola is a conic section created when a plane meets a double right circular cone at an angle that overlaps both cone halves. Along the conjugate axis, a hyperbola is symmetric. We will use parametric coordinates of the hyperbola in this question.





Complete step by step solution:The equation of the hyperbola is: $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
In parametric coordinates, the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 $ is represented by the equations $x=a sec\theta and y=b tan\theta $together. So, let the point on the hyperbola be$ (a sec\theta,b tan\theta).$
The equation of the tangent is:$ bxsec\theta-aytan\theta=ab$
Substituting x = 0, we get$ b=-ytan\theta$
so, $y=-bcot\theta = 1$
$b=-tan\theta$
Substituting y = 0, we get$ a=xsec\theta$
so, $x=acos\theta = 1$
$a=sec\theta$
As we know that the identity is $\sec^{2}\theta-tan^{2}\theta=1$
$a^{2}-(-b)^{2}=1$
$a^{2}-b^{2}=1$
$x^{2}-y^{2}=1$
Thus, a tangent to a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ intercepts a length of unity from each of the coordinate axes, then the point (a, b) lies on the rectangular hyperbola $x^{2}-y^{2}=1.$



Option ‘B’ is correct

Additional Information: When we replace $b^{2}$ in an ellipse by $(-b)^{2}$, we get a hyperbola.



Note: In a hyperbola, there are two axes. The transverse axis is the line that passes through the foci. The conjugate axis is the line through the centre that is also perpendicular to the transverse axis.