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

Answer 1

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

$4y^2=x^2+1$

$\Rightarrow –x^2+4y^2=1$

$\Rightarrow -\frac{x^2}{1^2}+\frac{y^2}{{\left(\frac{1}{2}\right)}^2}=1$

$a=1,b=\frac{1}{2}$

Let, tangent to the curve is at point $(x_1,y_1)$.

$\therefore 4\times 2y.\frac{dy}{dx}=2x$

$\Rightarrow \frac{dy}{dx}=\frac{2x_1}{8y_1}=\frac{x_1}{4y_1}$

$\therefore E{q}^n$ of tangent: $y=mx+c$

$\Rightarrow y=\frac{x_1}{4y_1}\cdot x+c$

$\Rightarrow y_1=\frac{x_1{y}_1}{4y_1}+c$

$\Rightarrow c=y_1=\frac{x_1^2}{4y_1}$

$=\frac{4y_1^2–x_1^2}{4y_1}=\frac{1}{4y_1}$

$\Rightarrow y=\frac{x_1}{4y_1}x+\frac{1}{4y_1}$

$\Rightarrow 4y_{1}y=x_{1}x+1\dots ..\left(I\right)$

Intersects $x$ axis at $(\frac{-1}{x_1},0)$

And $y$ axis at $=(0,\frac{1}{4y_1})$

$h=\frac{-1}{2x_1}$

$x_1=\frac{-1}{2h}$

$y_1=\frac{1}{8k}$

Midpoint : $(\frac{-1}{2x_1},\frac{1}{8y_1}):(h,k)$

$4y_1^2\ne x_1^2+1$

$\Rightarrow 4{\left(\frac{1}{8k}\right)}^2={\left(\frac{-1}{2h}\right)}^2+1$

$\Rightarrow \frac{1}{16k^2}=\frac{1}{4h^2}+1$

$\Rightarrow 1=\frac{16k^2}{4h^2}+16k^2$

$h^2=4k^2+16h^{2}k$.

$x^2–4y^2–16x^2{y}^2=0$

This is the required equation.