Skip to main content
Back to Conic Sections
JEE Main 2018
Conic Sections
Hyperbola
Easy

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

Solution

1. Key Concepts and Formulas

This problem requires the application of two core concepts from conic sections:

  • Equation of a Chord with a Given Midpoint: For a general conic SAx2+Bxy+Cy2+Dx+Ey+F=0S \equiv Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, if (h,k)(h, k) is the midpoint of a chord, the equation of that chord is given by T=S1T = S_1.

    • TT is obtained by replacing x2xhx^2 \to xh, y2yky^2 \to yk, xy12(xk+yh)xy \to \frac{1}{2}(xk + yh), x12(x+h)x \to \frac{1}{2}(x+h), y12(y+k)y \to \frac{1}{2}(y+k), and the constant term remains unchanged.
    • S1S_1 is obtained by substituting the midpoint coordinates (h,k)(h, k) into the equation of the conic, i.e., S1=Ah2+Bhk+Ck2+Dh+Ek+FS_1 = Ah^2 + Bhk + Ck^2 + Dh + Ek + F.
  • Condition for Tangency of a Line to a Parabola: A straight line with equation y=mx+cy = mx + c is tangent to the standard parabola y2=4axy^2 = 4ax if and only if c=amc = \frac{a}{m}.


2. Step-by-Step Solution

Step 1: Determine the Equation of the Chord of the Hyperbola

We are given the hyperbola x2y2=4x^2 - y^2 = 4. We can rewrite this as Sx2y24=0S \equiv x^2 - y^2 - 4 = 0. Let (h,k)(h, k) be the midpoint of a chord of this hyperbola. Our goal is to find the locus of (h,k)(h, k).

The equation of a chord whose midpoint is (h,k)(h, k) is given by the formula T=S1T = S_1.

  • Calculate TT (Tangent at (h,k)(h,k) form): To find TT, we apply the substitution rules to the hyperbola equation x2y24=0x^2 - y^2 - 4 = 0. Replace x2x^2 with xhxh and y2y^2 with ykyk. The constant term remains unchanged. T=xhyk4T = xh - yk - 4

  • Calculate S1S_1 (Value of S at (h,k)(h,k)): To find S1S_1, we substitute the coordinates of the midpoint (h,k)(h, k) into the equation of the hyperbola. S1=h2k24S_1 = h^2 - k^2 - 4

  • Equate T=S1T = S_1 to get the Chord Equation: Now, we set TT equal to S1S_1: xhyk4=h2k24xh - yk - 4 = h^2 - k^2 - 4 The constant term 4-4 cancels out from both sides: xhyk=h2k2xh - yk = h^2 - k^2

    Explanation: The T=S1T=S_1 formula is a powerful shortcut. It essentially represents the equation of a line that passes through the midpoint (h,k)(h,k) and is perpendicular to the diameter passing through (h,k)(h,k). This is precisely the definition of a chord with a given midpoint.

    To prepare for the tangency condition, we need to express this chord equation in the slope-intercept form y=mx+cy = mx + c. Rearranging the equation: yk=xh(h2k2)yk = xh - (h^2 - k^2) Assuming k0k \neq 0 (if k=0k=0, the chord is vertical, which cannot be tangent to y2=8xy^2=8x unless h=0h=0 which is a degenerate case), we can divide by kk: y=(hk)x(h2k2k)y = \left(\frac{h}{k}\right)x - \left(\frac{h^2 - k^2}{k}\right) By comparing this with y=mx+cy = mx + c, we identify the slope mm and y-intercept cc of the chord: m=hkandc=h2k2km = \frac{h}{k} \quad \text{and} \quad c = -\frac{h^2 - k^2}{k}

Step 2: Apply the Tangency Condition to the Parabola

The problem states that these chords touch the parabola y2=8xy^2 = 8x. To use the tangency condition c=a/mc = a/m, we first need to identify the parameter aa for the given parabola. We compare y2=8xy^2 = 8x with the standard form y2=4axy^2 = 4ax. From this comparison, we see that 4a=84a = 8, which implies a=2a = 2.

Now, we substitute the values of mm, cc (from Step 1), and aa into the tangency condition c=a/mc = a/m: h2k2k=2h/k-\frac{h^2 - k^2}{k} = \frac{2}{h/k}

Step 3: Simplify and Determine the Locus

Now, we simplify the equation obtained in Step 2: h2k2k=2kh-\frac{h^2 - k^2}{k} = \frac{2k}{h} To eliminate the denominators, we multiply both sides by hkhk (assuming h0h \neq 0 and k0k \neq 0): h(h2k2)=2k2-h(h^2 - k^2) = 2k^2 Distribute the h-h on the left side: h3+hk2=2k2-h^3 + hk^2 = 2k^2 We want to find the locus, which means an equation relating hh and kk. Let's rearrange the terms to isolate k2k^2: hk22k2=h3hk^2 - 2k^2 = h^3 Factor out k2k^2 from the terms on the left side: k2(h2)=h3k^2(h - 2) = h^3

Finally, to express the locus, we replace the general midpoint coordinates (h,k)(h, k) with the standard variables (x,y)(x, y): y2(x2)=x3y^2(x - 2) = x^3

This equation represents the locus of the midpoints of the chords.

Practice More Conic Sections Questions

View All Questions