Question:hard

If a tangent drawn to the parabola \( y^{2}=16x \) meets the curve \( xy=4 \) at the points P and Q, then the locus of midpoint of PQ is

Show Hint

The combination of a slope-form line equation and the midpoint-chord formula \( T = S_1 \) is the standard way to solve dual-curve intersection locus problems without doing actual coordinate factorization.
Updated On: Jun 7, 2026
  • \( y^{2}=2x \)
  • \( y^{2}+2x=0 \)
  • \( y^{2}=4x \)
  • \( y^{2}+4x=0 \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Write a tangent to the parabola.
For $y^2=16x$, we have $4a=16$, so $a=4$. A slope-form tangent is $y=mx+\frac{4}{m}$, which rearranges to \[ m^2x-my+4=0\quad\cdots(1) \]
Step 2: Write the chord of the hyperbola with midpoint $(h,k)$.
For $xy=4$, the chord whose midpoint is $(h,k)$ comes from $T=S_1$: \[ kx+hy=2hk\quad\cdots(2) \]
Step 3: Match the two lines.
Equations (1) and (2) are the same straight line, so their coefficients are proportional: \[ \frac{m^2}{k}=\frac{-m}{h}=\frac{4}{-2hk} \]
Step 4: Find $m$ from the first pair.
\[ \frac{m^2}{k}=\frac{-m}{h}\implies m=-\frac{k}{h} \]
Step 5: Use the last ratio.
\[ \frac{-m}{h}=\frac{4}{-2hk}=\frac{-2}{hk}\implies m=\frac{2}{k} \]
Step 6: Combine and read the locus.
Setting $-\frac{k}{h}=\frac{2}{k}$ gives $-k^2=2h$, that is $k^2+2h=0$. Replacing $(h,k)$ by $(x,y)$: \[ \boxed{y^2+2x=0} \] (A quick check: the tangent $y=x+4$ meets $xy=4$ at points whose midpoint is $(-2,2)$, and $2^2+2(-2)=0$, which fits.)
Was this answer helpful?
0