If the tangents drawn to the hyperbola 4 y^2=x^2+ 1 intersect the...

Equation of hyperbola is : $$\begin{aligned}& 4 y²=x²+1 \Rightarrow-x²+4 y²=1 \\ \Rightarrow &-\frac{x^2}{1^2}+\frac{y^2}{\left(\frac{1}{2}\right)^2}=1 \\ \therefore & a=1, b=\frac{1}{2} \end{aligned}$$ Now, tangent to the curve at point \left(x_1, y_1\right)$is given by.$$\begin{aligned}&4 \times2 y_1 \frac{d y}{d x}=2 x_1 \\ &\Rightarrow \frac{d y}{d x}=\frac{2 x_1}{8 y_1}=\frac{x_1}{4 y_1} \end{aligned}$$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$$\begin{aligned}&\therefore y_1=\frac{x_1 x_1}{4 y_1}+ c \\ &\Rightarrow C=\frac{4 y_1^2-x_1^2}{4 y_1}=\frac{1}{4 y_1} \end{aligned}$Therefore, $y=\frac{x_1}{4 y_1} x+\frac{1}{4 y_1}$ \Rightarrow 4 y_1 y=x_1 x+1 which intersects $ x$ axis at $A $\left(\frac{-1}{x_1},0\right)$ and $y$ axis at $B \left(0, \frac{1}{4 y_1}\right)$ Let midpoint of $A B$ is $(h, k)\begin{aligned}&\therefore h=\frac{-1}{2 x_1}\\ &\Rightarrow x_1=\frac{-1}{2 h}& y_1=\frac{1}{8 k} \end{aligned}$Thus, 4\left(\frac{1}{8 k}\r right)^2=\left(\frac{-1}{2 h}\right)^2+1\begin{aligned}&\Rightarrow \frac{1}{16 k²}=\frac{1}{4 h²}+1 \\ &\Rightarrow 1=\frac{16 k²}{4 h²}+16 k² \quad \Rightarrow h²=4 k²+16 h² k . \end{aligned}$$ So, required equation is x²-4 y²-16 x² y²=0