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 :

• A. \(x^3(x-2) = y^2\)
• B. \(x^2(x-2) = y^3\)
• C. \(y^2(x-2) = x^3\)
• D. \(y^3(x-2) = x^2\)

Hint:

The formula \(T=S_1\) is a very useful tool for finding the equation of a chord of any conic section when its midpoint is known. Combining this with the condition of tangency (\(c=a/m\) for parabola, \(c^2=a^2m^2 \pm b^2\) for ellipse/hyperbola) is a standard method for solving locus problems of this type.

Answer 1

The Correct Option is C

Step 1: Equation of the Chord with a Given Midpoint
Let the midpoint of a chord of the hyperbola \(x^2-y^2=4\) be \((h, k)\). The equation of the chord is given by the formula \(T=S_1\), where \(T = xh - yk - 4\) and \(S_1 = h^2 - k^2 - 4\). \[ xh - yk - 4 = h^2 - k^2 - 4 \] \[ xh - yk = h^2 - k^2 \] This is the equation of the chord.
Step 2: Condition of Tangency
This chord touches the parabola \(y^2 = 8x\). Let's rewrite the chord's equation in the form \(y=mx+c\). \[ yk = xh - (h^2 - k^2) \implies y = \frac{h}{k}x - \frac{h^2 - k^2}{k} \] So, the slope is \(m = \frac{h}{k}\) and the y-intercept is \(c = -\frac{h^2-k^2}{k}\). The condition for a line \(y=mx+c\) to be tangent to the parabola \(y^2 = 4ax\) is \(c = \frac{a}{m}\). For the parabola \(y^2 = 8x\), we have \(4a=8\), so \(a=2\).
Step 3: Derive the Locus
Applying the condition of tangency: \[ c = \frac{a}{m} \implies -\frac{h^2-k^2}{k} = \frac{2}{h/k} = \frac{2k}{h} \] \[ -h(h^2-k^2) = 2k^2 \] \[ -h^3 + hk^2 = 2k^2 \] \[ hk^2 - 2k^2 = h^3 \] \[ k^2(h-2) = h^3 \] Step 4: Final Answer
To find the locus, we replace the general point \((h, k)\) with \((x, y)\). \[ y^2(x-2) = x^3 \] This is the required locus of the midpoints.