Conic Sections: JEE Main Mathematics Question with Solution
Let a tangent to the curve y2=24x meet the curve xy=2 at the points A and B. Then the mid points of such line segments AB lie on a parabola with the
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Hint 1 of 4
Let the midpoint of the chord AB on the hyperbola xy=2 be (h,k). What is the equation of the chord AB using T=S1?
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Correct answer
The locus of the midpoints is the parabola y2=−3x, which has directrix x=43, or 4x=3.
Option analysis
Why each option works or fails
A · Length of latus rectum 23
The student misidentifies the coefficient of x in the locus equation y2=−3x or calculates the latus rectum as 4a/2 instead of 4a=3. For a standard parabola y2=−4ax, the length of the latus rectum is the absolute value of the coefficient of x, which is 3, not 23.
B · directrix 4x=−3
The student finds y2=−3x correctly but misapplies the directrix formula as x=−a instead of x=a for a parabola opening to the left. For a parabola opening to the left, y2=−4ax with a=43, the directrix lies on the positive side: x=a=43, which simplifies to 4x=3.
C · length of latus rectum 2
The student makes an algebraic error when applying the condition of tangency (c=a/m), obtaining an incorrect coefficient such as y2=−2x. Substitute the midpoint chord equation xy1+yx1=2x1y1 into slope-intercept form and equate the intercept with c=m6 to ensure the correct coefficient of 3.
D · directrix 4x=3
None. The student correctly applies T=S1 for the hyperbola xy=2, writes the line in slope-intercept form, applies the tangency condition to y2=24x, and determines the directrix of the resulting parabola. Correctly identify the locus as y2=−3x, which corresponds to y2=−4ax with a=43, giving the directrix x=43 or 4x=3.
Reviewed route
Solution
StepWorking
01given
Parabola C1:y2=24x, rectangular hyperbola C2:xy=2. Line segment AB is formed by a tangent to C1 intersecting C2 at points A and B. Let M(h,k) be the midpoint of chord AB.
02approach
Write the tangent to y2=24x in parametric form. Then write the equation of chord AB of xy=2 having midpoint (h,k) using T=S1. Compare coefficients of both equations of the same line AB to eliminate the parameter t and obtain the locus of (h,k).
03execute
For y2=4ax=24x, we have a=6. The tangent to C1 at (6t2,12t) is ty=x+6t2, which can be rewritten as x−ty+6t2=0.
For C2:xy−2=0, the chord with midpoint (h,k) is given by T=S1:
21(xk+yh)−2=hk−2⟹kx+hy=2hk⟹hx+ky=2, which is hx+ky−2=0.
Comparing x−ty=−6t2 and hx+ky=2:
1/h1=1/k−t=2−6t2⟹h=−kt=−3t2.
From h=−3t2 and −kt=−3t2⟹k=3t, so t=3k.
Substitute t into h=−3t2:
h=−3(3k)2=−39k2=−3k2⟹k2=−3h.
Thus, the locus of the midpoint (h,k) is the parabola y2=−3x.
✓verify
The locus is y2=−3x, which is in standard form y2=−4Ax where 4A=3⟹A=3/4.
The length of the latus rectum is 4A=3.
The directrix of y2=−4Ax is x=A, so x=43⟹4x=3. This matches option (3).
Hints that build this answer step by step
Let the midpoint of the chord AB on the hyperbola xy=2 be (h,k). What is the equation of the chord AB using T=S1?
kx+hy=2hk
Expressing this chord in slope-intercept form y=mx+c, what are the slope m and y-intercept c in terms of h and k?
m=−hk and c=2k
The line y=mx+c is tangent to the parabola y2=24x (where 4a=24, so a=6). Applying the condition of tangency c=ma, what locus equation relating h and k is obtained?
k2=−3h, so the locus is y2=−3x
For the parabola y2=−3x, what is the equation of its directrix?
Why is the chord equation T = S_1 written as (x/h) + (y/k) = 2?
For the curve xy - c^2 = 0, T is (x k + y h)/2 - c^2 and S_1 is hk - c^2. Equating T = S_1 gives (xk + yh)/2 = hk, which upon dividing by hk yields x/h + y/k = 2.