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 :

Question

If a function is increasing, then its derivative is:

Answer 1

To find the locus of the midpoint of the tangent points on the hyperbola given by the equation 4y^2 = x^2 + 1, we first consider its standard form. Rewriting, we have:

\frac{x^2}{4} - \frac{y^2}{1} = -1

This represents a hyperbola. The equation of a tangent to this hyperbola at point (x_1, y_1) will be given by:

\frac{xx_1}{4} - yy_1 = 1

This line intersects the x-axis (y = 0) at the point A, and the y-axis (x = 0) at the point B. Let's determine these intercepts:

For the x-axis intercept, set y = 0:
\frac{xx_1}{4} = 1

Solving for x, we get x = \frac{4}{x_1}.
For the y-axis intercept, set x = 0:
-yy_1 = 1

Solving for y, we get y = -\frac{1}{y_1}.

The midpoint (h, k) of the segment AB, where A\left(\frac{4}{x_1}, 0\right) and B(0, -\frac{1}{y_1}), is:

h = \frac{\frac{4}{x_1} + 0}{2} = \frac{2}{x_1}, k = \frac{0 - \frac{1}{y_1}}{2} = -\frac{1}{2y_1}

Now, we need to find the relation between h and k. From the point (x_1, y_1) on the hyperbola 4y_1^2 = x_1^2 + 1, we have:

h = \frac{2}{x_1} \implies x_1 = \frac{2}{h} and k = -\frac{1}{2y_1} \implies y_1 = -\frac{1}{2k}

Substituting into the hyperbola's equation:

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

Simplify:

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

Cross-multiply and rearrange to obtain:

h^2 - 4k^2 - 16k^2h^2 = 0

Replacing h with x and k with y to express the locus:

The locus of the midpoint (x, y) is x^2 - 4y^2 - 16x^2y^2 = 0.

This matches the correct option given. Thus, the answer is:

Correct Answer: x^2 - 4y^2 - 16x^2y^2 = 0

Answer 2