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

Answer 1

Answer: (D)

We have,
$4y^{2}=x^{2}+1$
$\Rightarrow 4y^{2}-x^{2}=1$
Then,
$8y\frac{d y}{d x}=2x$
$\Rightarrow \frac{d y}{d x}=\frac{x}{4 y}$
This is the conjugate hyperbola.
Let the tangent to the curve is at point $\left(x_{1} , y_{1}\right)$ , then slope of tangent is
$\left(\frac{d y}{d x}\right)_{\left(x_{1} , y_{1}\right)}=\frac{x_{1}}{4 y_{1}}$
Then equation of tangent is
$y=mx+c$
$\Rightarrow y=\left(\frac{x_{1}}{4 y_{1}}\right)x+c$
$\Rightarrow c=y-\left(\frac{x_{1}}{4 y_{1}}\right)x$
$\Rightarrow c=\frac{4 y_{1}^{2} - x x_{1}}{4 y_{1}}$
$\Rightarrow c=\frac{1}{4 y_{1}}$
Hence, equation of tangent is
$y=\left(\frac{x_{1}}{4 y_{1}}\right)x+\left(\frac{1}{4 y_{1}}\right)$
$\Rightarrow 4y_{1}y=x_{1}x+1$
Putting $y=0$ , we get
$A\equiv \left(\frac{- 1}{x_{1}} , 0\right)$
And, putting $x=0$ , we get $B\equiv \left(0 , \frac{1}{4 y_{1}}\right)$
Let the mid-point of $AB$ be $\left(h , k\right)$ then
$h=\frac{- 1}{2 x_{1}}\Rightarrow x_{1}=\frac{- 1}{2 h}$
And, $k=\frac{1}{8 y_{1}}\Rightarrow y_{1}=\frac{1}{8 k}$
Satisfying $\left(x_{1} , y_{1}\right)$ in the equation of hyperbola, we get
$h^{2}-k^{2}-16h^{2}k^{2}=0$
Hence, required locus is $x^{2}-y^{2}-16x^{2}y^{2}=0$