If the tangents drawn to the hyperbola 4y^2 = | vedclass

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(D) The equation of the hyperbola is $4y^2 - x^2 = 1$.Let the point of tangency be $P(x_1, y_1)$. The equation of the tangent at $P$ is $4yy_1 - xx_1

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$x^2 - 4y^2 - 16x^2y^2 = 0$

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(D) The equation of the hyperbola is $4y^2 - x^2 = 1$.Let the point of tangency be $P(x_1, y_1)$. The equation of the tangent at $P$ is $4yy_1 - xx_1 = 1$.This tangent intersects the $x$-axis at $A$ where $y=0$,so $-xx_1 = 1 \Rightarrow x = -1/x_1$. Thus,$A = (-1/x_1, 0)$.This tangent intersects the $y$-axis at $B$ where $x=0$,so $4yy_1 = 1 \Rightarrow y = 1/(4y_1)$. Thus,$B = (0, 1/(4y_1))$.Let the midpoint of $AB$ be $(h, k)$.Then $h = -1/(2x_1) \Rightarrow x_1 = -1/(2h)$ and $k = 1/(8y_1) \Rightarrow y_1 = 1/(8k)$.Since $(x_1, y_1)$ lies on the hyperbola $4y_1^2 - x_1^2 = 1$,we substitute the values:$4(1/(8k))^2 - (-1/(2h))^2 = 1$$4/(64k^2) - 1/(4h^2) = 1$$1/(16k^2) - 1/(4h^2) = 1$Multiplying by $16h^2k^2$:$h^2 - 4k^2 = 16h^2k^2$Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - 4y^2 - 16x^2y^2 = 0$.

If the tangents drawn to the hyperbola $4y^2 = x^2 + 1$ intersect the coordinate axes at the distinct points $A$ and $B$,then the locus of the midpoint of $AB$ is

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