1.

Let the transverse axis ofa varying hyperbola be fixed with length of transverse axis being 2a. Then the locus of the point of contact of any tangent drawn to it from a fixed point on conjugate axis is

Answer»

<P>a PARABOLA
a circle
an ellipse
a hyperbola

Solution :Let the hyperbola be `(x^(2))/(a^(2)) -(y^(2))/(b^(2)) =1`. Any tangent to it is `(x)/(a) sec phi - (y)/(b) tan phi =1` (1)
at `Q (a sec phi, b tan theta)`
The tangent CUTS the axis of y at P. The coordinates of P are `(0,-b cot phi)`
As P is fixed `rArr b cot phi = lambda` (say) (2)
Now `x = a sec phi, y = b tan phi` (3)
ELIMINATING b and `phi` from (1),(2),(3), we get `(x^(2))/(a^(2)) -(y)/(lambda) =1`, which clearly is a parabola.


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