If the tangents drawn to the hyperbola 4y²=x²+1 intersect the co-ordinate axes at the distinct point

Q: Wait, why do we use the $T=0$ concept here?

The $T=0$ method is a powerful, universal tool in coordinate geometry for finding the equation of a tangent to any conic section at a point $(x_1, y_1)$. It saves you from the tedious process of differentiation and is the standard, elegant way to handle such problems in JEE Advanced.

Q: Why do we replace $(h, k)$ with $(x, y)$ at the end?

The locus is the set of all possible positions of the midpoint. By assuming the midpoint is $(h, k)$ and finding a relationship between $h$ and $k$, we are essentially describing the path it traces. Replacing $(h, k)$ with $(x, y)$ is just the standard convention to express the final equation in the Cartesian plane.

Q: Can we use parametric coordinates instead of $(x_1, y_1)$?

Absolutely! You could use the parametric form for a hyperbola, such as $x = \sec\theta$ and $y = \frac{1}{2}\tan\theta$. However, the algebraic method using $(x_1, y_1)$ is often more direct and less prone to trigonometric identity errors. Both paths lead to the same beautiful result.

Question

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 :

Options:

(A) $4x^2 - y^2 - 16x^2y^2 = 0$

(B) $4x^2 - y^2 + 16x^2y^2 = 0$

(C) $x^2 - 4y^2 + 16x^2y^2 = 0$

(D) $x^2 - 4y^2 - 16x^2y^2 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to a beautiful problem in coordinate geometry. Today, we are not just solving an equation; we are tracing the path of a point that dances along the tangents of a hyperbola.

Let us begin by looking at our curve: $4y^2 = x^2 + 1$. To truly understand its nature, we must bring it into its standard form.

By rearranging the terms, we get $4y^2 - x^2 = 1$. Dividing by $1$, we see the structure:

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

This is a vertical hyperbola, opening along the $y$-axis. It is the stage upon which our drama unfolds.

### The Tangent's Dance

Imagine a point $P(x_1, y_1)$ resting on this hyperbola. Because it is a resident of this curve, it must obey the law: $4y_1^2 - x_1^2 = 1$.

Now, we draw a tangent at this point. Instead of using calculus, we invoke the elegant $T=0$ concept. For our hyperbola, the equation of the tangent at $(x_1, y_1)$ is simply:

$$
4yy_1 - xx_1 = 1
$$

This line is our protagonist. It cuts through the coordinate axes at two distinct points, $A$ and $B$.

To find $A$, the $x$-intercept, we set $y=0$, yielding $A\left(-\frac{1}{x_1}, 0\right)$. To find $B$, the $y$-intercept, we set $x=0$, yielding $B\left(0, \frac{1}{4y_1}\right)$.

### The Midpoint's Path

The question asks for the locus of the midpoint $M(h, k)$ of the segment $AB$. Using the midpoint formula, we find:

$$
h = \frac{-\frac{1}{x_1} + 0}{2} = -\frac{1}{2x_1}
$$

$$
k = \frac{0 + \frac{1}{4y_1}}{2} = \frac{1}{8y_1}
$$

We are almost there! We have the coordinates of our midpoint in terms of the point of tangency. To find the locus, we must eliminate the parameters $x_1$ and $y_1$.

Rearranging our equations, we get $x_1 = -\frac{1}{2h}$ and $y_1 = \frac{1}{8k}$.

### The Final Revelation

Now, we return to the condition that defined our point $P$: $4y_1^2 - x_1^2 = 1$. Substituting our expressions for $x_1$ and $y_1$, we get:

$$
4\left(\frac{1}{8k}\right)^2 - \left(-\frac{1}{2h}\right)^2 = 1
$$

Expanding this, we have $4\left(\frac{1}{64k^2}\right) - \frac{1}{4h^2} = 1$, which simplifies to:

$$
\frac{1}{16k^2} - \frac{1}{4h^2} = 1
$$

Multiplying by $16h^2k^2$ to clear the denominators, we arrive at $h^2 - 4k^2 = 16h^2k^2$.

Replacing $(h, k)$ with $(x, y)$, we find the final locus:

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

This matches our fourth option. You see, the complexity melts away when you follow the logic step by step. Keep practicing, and you will master these beautiful curves.

Animated Solution for Mathematics - Conic Sections: 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 AB is :

Select Answer:

Visualized Solution

Standardizing the Hyperbola

Defining the Point of Tangency

Equation of Tangent at $P$

Finding the x-intercept $A$

Finding the y-intercept $B$

Midpoint of Segment $A B$

Applying the Midpoint Formula

Isolating $x_{1}$ and $y_{1}$

Substituting into the Curve Equation

Expanding the Equation

Simplifying the Locus Equation

Final Locus Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Dance

The Midpoint's Path

The Final Revelation

Similar Questions

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

From the origin chords are drawn to the circle $(x - 1)^{2} + y^{2} = 1$ . The equation of the locus of the mid-points of these chords is .........

A normal to the hyperbola, $4 x^{2} - 9 y^{2} = 36$ meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the locus of P is :-

Consider the family of circles $x^{2} + y^{2} = r^{2}, 2 < r < 5$ . If in the first quadrant, the common tangent to a circle of this family and the ellipse $4 x^{2} + 25 y^{2} = 100$ meets the co-ordinate axes at $A$ and $B$ , then find the equation of the locus of the mid-point of $A B$ .

Tangents are drawn to the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ , parallel to the straight line $2 x - y = 1$ . The points of contact of the tangents on the hyperbola are

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :

The locus of a point $P (α, β)$ moving under the condition that the line $y = α x + β$ is a tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is

A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point: