The locus of the mid points of the chords of the hyperbola x² - y² = 4, which touch the parabola y²

Q: Wait, why do we use $T=S_1$ instead of just finding the slope and using point-slope form?

Using $T=S_1$ is the most elegant and direct path in coordinate geometry. While you could find the slope by calculating the derivative at the intersection points, $T=S_1$ is a powerful identity specifically designed to define a chord by its midpoint. It saves you from the algebraic nightmare of solving for intersection points!

Q: How do I know if the line is tangent to the parabola or the hyperbola?

The problem explicitly states the chord is a tangent to the parabola $y^2 = 8x$. The hyperbola $x^2 - y^2 = 4$ is simply the curve that 'hosts' the chord. We use the hyperbola to define the line and the parabola to constrain it.

Q: Why is the condition for tangency $c = a/m$?

For a parabola $y^2 = 4ax$, any line $y = mx + c$ is tangent if and only if $c = a/m$. You can derive this by substituting $y = mx + c$ into $y^2 = 4ax$ and setting the discriminant of the resulting quadratic to zero. It is a fundamental tool in your JEE arsenal!

Q: What happens if $k=0$ in my final equation?

If $k=0$, the midpoint lies on the x-axis. Looking at our final equation $y^2(x-2) = x^3$, if $y=0$, then $x$ must be $0$. This corresponds to the vertex of the parabola, which is a valid point in our locus. Always check your boundaries!

Q: Is there any other way to solve this without $T=S_1$?

Yes, you could parameterize the chord using two points on the hyperbola, but that would involve complex trigonometric or hyperbolic functions. In JEE Advanced, the 'smart' way is usually the 'fast' way. $T=S_1$ is the gold standard for midpoint problems.

Question

The locus of the mid points of the chords of the hyperbola $x^2 - y^2 = 4$, which touch the parabola $y^2 = 8x$, is :

Options:

(A) $y^3(x - 2) = x^2$

(B) $x^3(x - 2) = y^2$

(C) $y^2(x - 2) = x^3$

(D) $x^2(x - 2) = y^3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast coordinate plane. On one side, you see the elegant, sweeping branches of the hyperbola $x^2 - y^2 = 4$. On the other, the focused, parabolic curve of $y^2 = 8x$.

We are looking for the path—the locus—traced by the midpoints of all possible chords of the hyperbola that happen to be tangent to the parabola.

### Defining the Chord

When we talk about a chord of a hyperbola, we are talking about a line segment connecting two points on the curve. To define this line using its midpoint $(h, k)$, we utilize the identity $T=S_1$.

For our hyperbola $S: x^2 - y^2 - 4 = 0$, the expression $T$ is the tangent-like form $xh - yk - 4$, and $S_1$ is the value of the conic at the point $(h, k)$, which is $h^2 - k^2 - 4$. By setting $T=S_1$, we obtain:

$$xh - yk - 4 = h^2 - k^2 - 4$$

Simplifying this, we find the equation of our chord:

$$xh - yk = h^2 - k^2$$

### The Tangency Condition

Now, we require this line to be a tangent to the parabola $y^2 = 8x$. To facilitate this, we rewrite our chord equation in the slope-intercept form $y = mx + c$:

$$yk = xh - (h^2 - k^2) \implies y = \left( \frac{h}{k} \right)x - \frac{h^2 - k^2}{k}$$

Here, our slope $m$ is $\frac{h}{k}$ and our intercept $c$ is $-\frac{h^2 - k^2}{k}$. For a line to be tangent to a parabola $y^2 = 4ax$, it must satisfy the condition $c = \frac{a}{m}$.

Given $4a = 8$, we have $a = 2$. Substituting our values into the tangency condition $c = \frac{a}{m}$, we get:

$$-\frac{h^2 - k^2}{k} = \frac{2}{h/k}$$

### The Final Synthesis

We now simplify the algebraic relationship connecting the coordinates of the midpoint $(h, k)$ to the geometry of the parabola:

$$-\frac{h^2 - k^2}{k} = \frac{2k}{h}$$

Cross-multiplying yields:

$$-h(h^2 - k^2) = 2k^2$$

$$-h^3 + hk^2 = 2k^2$$

Rearranging to isolate the terms involving $k^2$, we obtain:

$$hk^2 - 2k^2 = h^3$$

$$k^2(h - 2) = h^3$$

Finally, replacing $(h, k)$ with $(x, y)$ to express the locus in standard coordinates, we arrive at the final equation:

**$$y^2(x - 2) = x^3$$**

This is the path traced by the midpoints. By applying the properties of conics, we have mapped a complex geometric relationship into a single, clean algebraic expression.

Animated Solution for Mathematics - Conic Sections: The locus of the mid points of the chords of the hyperbola $x^{2} - y^{2} = 4$ , which touch the parabola $y^{2} = 8 x$ , is :

Select Answer:

Visualized Solution

Visualize the Geometric Setup

Define the Midpoint $(h, k)$

Apply $T = S_{1}$ to the Hyperbola

Formulate the Chord Equation

Convert to Slope-Intercept Form

Identify Tangency Parameters

The Condition for Tangency

Substitute and Set Up Equation

Cross-Multiply and Simplify

Group Terms to Find the Locus

Final Locus Equation

The Sigma Insight: Chord in terms of Midpoint

Analyzing the Setup

Defining the Chord

The Tangency Condition

The Final Synthesis

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the mid points of the chords of the hyperbola x2−y2=4, which touch the parabola y2=8x, is :