If the tangents drawn to the hyperbola $4 y^2=x^2+$ 1 intersect the co-ordinate axes at the distinct points…
If the tangents drawn to the hyperbola $4 y^2=x^2+$ 1 intersect the co-ordinate axes at the distinct points $A$ and $B$, then the locus of the mid point of $A B$ is
$x^2-4 y^2+16 x^2 y^2=0$
$4 x^2-y^2+16 x^2 y^2=0$
$4 x^2-y^2-16 x^2 y^2=0$
$x^2-4 y^2-16 x^2 y^2=0$
Solution
Equation of hyperbola is :
$
\begin{aligned}
& 4 y^2=x^2+1 \Rightarrow-x^2+4 y^2=1 \\
\Rightarrow &-\frac{x^2}{1^2}+\frac{y^2}{\left(\frac{1}{2}\right)^2}=1 \\
\therefore & a=1, b=\frac{1}{2}
\end{aligned}
$
Now, tangent to the curve at point $\left(x_1, y_1\right)$ is given by.
$
\begin{aligned}
&4 \times 2 y_1 \frac{d y}{d x}=2 x_1 \\
&\Rightarrow \frac{d y}{d x}=\frac{2 x_1}{8 y_1}=\frac{x_1}{4 y_1}
\end{aligned}
$
Equation of tangent at $\left(x_1, y_1\right)$ is
$
\begin{aligned}
&y=m x+c \\
&\Rightarrow y=\frac{x_1}{4 y_1} \cdot x+c
\end{aligned}
$
As tangent passes through $x_1, y_1$
$
\begin{aligned}
&\therefore y_1=\frac{x_1 x_1}{4 y_1}+c \\
&\Rightarrow C=\frac{4 y_1^2-x_1^2}{4 y_1}=\frac{1}{4 y_1}
\end{aligned}
$
Therefore, $y=\frac{x_1}{4 y_1} x+\frac{1}{4 y_1}$
$
\Rightarrow 4 y_1 y=x_1 x+1
$
which intersects $x$ axis at $A\left(\frac{-1}{x_1}, 0\right)$ and $y$ axis at $B\left(0, \frac{1}{4 y_1}\right)$
Let midpoint of $A B$ is $(h, k)$
$
\begin{aligned}
&\therefore h=\frac{-1}{2 x_1} \\
&\Rightarrow x_1=\frac{-1}{2 h} \& y_1=\frac{1}{8 k}
\end{aligned}
$
Thus, $4\left(\frac{1}{8 k}\right)^2=\left(\frac{-1}{2 h}\right)^2+1$
$
\begin{aligned}
&\Rightarrow \frac{1}{16 k^2}=\frac{1}{4 h^2}+1 \\
&\Rightarrow 1=\frac{16 k^2}{4 h^2}+16 k^2 \quad \Rightarrow h^2=4 k^2+16 h^2 k .
\end{aligned}
$
So, required equation is
$
x^2-4 y^2-16 x^2 y^2=0
$