Question

If the tangents drawn to the hyperbola $4y^2=x^2+1$ intersect the co-ordinate axes at the distinct points $A$ and $B$, then the locus of the mid point of $AB$ is :

• A. $x^2-4y^2+16x^2{y}^2=0$
• B. $x^2-4y^2-16x^2{y}^2=0$
• C. $4x^2-y^2+16x^2{y}^2=0$
• D. $4x^2-y^2-16x^2{y}^2=0$

Answer 1

The correct option is B $x^2-4y^2-16x^2{y}^2=0$

Any point on the hyperbola will be,
$(\tan \theta ,\frac{\sec \theta }{2})$
Equation of tangent, $T=0$
$4y\frac{\sec \theta }{2}=x\tan \theta +1\Rightarrow 2y\sec \theta =x\tan \theta +1$
Given that tangents drawn to the hyperbola intersect the co-ordinate axes at the distinct points $A$ and $B$
Thus the coordinates of $A$
$y=0,x=-\cot \theta$
And the coordinates of $B$
$x=0,y=\frac{\cos \theta }{2}$
Let the midpoint of $AB=(h,k)$
$(h,k)=(\frac{-\cot \theta +0}{2},\frac{0+\frac{\cos \theta }{2}}{2})\Rightarrow \frac{-1}{2h}=\tan \theta ,\frac{1}{4k}=\sec \theta \Rightarrow {\left(\frac{1}{4k}\right)}^2-{\left(\frac{-1}{2h}\right)}^2=1\Rightarrow 4h^2-16k^2=64k^2{h}^2$
Hence the locus of the midpoint will be,
$x^2-4y^2-16x^2{y}^2=0$