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,
$8y\frac{dy}{dx}=2x$
$\Rightarrow \frac{dy}{dx}=\frac{x}{4y}$
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
${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{x_1}{4y_1}$
Then equation of tangent is
$y=mx+c$
$\Rightarrow y=\left(\frac{x_1}{4y_1}\right)x+c$
$\Rightarrow c=y-\left(\frac{x_1}{4y_1}\right)x$
$\Rightarrow c=\frac{4y_1^2-x{x}_1}{4y_1}$
$\Rightarrow c=\frac{1}{4y_1}$
Hence, equation of tangent is
$y=\left(\frac{x_1}{4y_1}\right)x+\left(\frac{1}{4y_1}\right)$
$\Rightarrow 4y_{1}y=x_{1}x+1$
Putting $y=0$ , we get
$A\equiv \left(\frac{-1}{x_1},0\right)$
And, putting $x=0$ , we get $B\equiv \left(0,\frac{1}{4y_1}\right)$
Let the mid-point of $AB$ be $\left(h,k\right)$ then
$h=\frac{-1}{2x_1}\Rightarrow x_1=\frac{-1}{2h}$
And, $k=\frac{1}{8y_1}\Rightarrow y_1=\frac{1}{8k}$
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 $x^2-y^2-16x^2{y}^2=0$