If the tangents drawn to the hyperbola 4y2 = x2 + 1 intersect the co-ordinate axes at the distinct points A and B then the locus of the mid point of AB is :
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Official solution
Equation of hyperbola is :
4y2 = x2 + 1 \[ \Rightarrow \] - x2 + 4y2 = 1
\[ \Rightarrow \] \[\,\,\,\] - \[{{{x^2}} \over {{1^2}}}\] + \[{{{y^2}} \over {{{\left( {{1 \over 2}} \right)}^2}}}\] = 1
\[ \therefore \] a = 1, b = \[{1 \over 2}\]
Now, tangent to the curve at point (x1, y1) is given by
4 \[ \times \] 2y1\[{{dy} \over {dx}}\] = 2x1
\[ \Rightarrow \] \[{{dy} \over {dx}}\] = \[{{2{x_1}} \over {8{y_1}}}\] = \[{{{x_1}} \over {4{y_1}}}\]
Equation of tangent at (x1, y1) is
y = mx + c
\[ \Rightarrow \] y = \[{{{x_1}} \over {4{y_1}}}\] . x + c
As tangent passes through (x1, y1)
\[ \therefore \] y1 = \[{{{x_1}{x_1}} \over {4{y_1}}} + c\]
\[ \Rightarrow \] C = \[{{4y_1^2 - x_1^2} \over {4{y_1}}}\] = \[{1 \over {4{y_1}}}\]
Therefore, y = \[{{{x_1}} \over {4{y_1}}}x + {1 \over {4{y_1}}}\]
\[ \Rightarrow \] 4y1y = x1x + 1
which intersects x axis at A \[\left( {{{ - 1} \over {{x_1}}},0} \right)\] and y axis at \[B\left( {0,{1 \over {4{y_1}}}} \right)\]
Let midpoint of AB is (h, k)
\[ \therefore \] h = \[{{ - 1} \over {2{x_1}}}\]
\[ \Rightarrow \] x1 ={{ - 1} \over {2h}}\[ \& y1 = \]{1 \over {8k}}Thus, 4(8k1)2=(2h−1)2+ 1⇒\[\,\,\,\] \[{1 \over {16{k^2}}}\] = \[{1 \over {4{h^2}}}\] + 1
\[ \Rightarrow \] \[\,\,\,\] 1 = \[{{16{k^2}} \over {4{h^2}}}\] + 16k2
\[ \Rightarrow \] \,\,\,\[ h2 = 4k2 + 16h2 k.
So, required equation is
x2 \]-4y2−