Question

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

• A. \( y^{2}=2x \)
• B. \( y^{2}+2x=0 \)
• C. \( y^{2}=4x \)
• D. \( y^{2}+4x=0 \)

Hint:

Instead of writing the tangent in slope form, use the parametric tangent to the parabola at a general point t: ty = x + 4t^2. Substituting this into xy = 4 gives a quadratic in x whose sum of roots locates the midpoint directly, without needing a chord-midpoint formula for the hyperbola.

Answer 1

The Correct Option is B

Concept:

• The equation of any tangent to the parabola \( y^2 = 4ax \) in slope form is \( y = mx + \frac{a}{m} \).

• The equation of a chord of a hyperbola \( xy = c^2 \) with a given midpoint \( (h,k) \) is represented by \( T = S_1 \), which expands to: \[ \frac{xk + yh}{2} = hk \implies xk + yh = 2hk \]

Step 1: Writing the tangent line property.
For the parabola \( y^2 = 16x \), \( 4a = 16 \implies a = 4 \). The tangent equation in slope form is: \[ y = mx + \frac{4}{m} \implies m^2x - my + 4 = 0 \quad \cdots (1) \]

Step 2: Writing the chord equation for the hyperbola using midpoint \( (h,k) \).
The line serves as a chord for \( xy = 4 \) with midpoint \( (h,k) \). Using \( T = S_1 \): \[ x k + y h = 2hk \implies kx + hy - 2hk = 0 \quad \cdots (2) \]

Step 3: Comparing coefficients to eliminate slope parameter \( m \).
Since equation (1) and equation (2) represent the exact same line, their coefficient ratios must be identical: \[ \frac{m^2}{k} = \frac{-m}{h} = \frac{4}{-2hk} \] From the first two terms: \[ \frac{m^2}{-m} = \frac{k}{h} \implies -m = \frac{k}{h} \implies m = -\frac{k}{h} \] Now substitute this expression for \( m \) into the relationship with the constant term: \[ \frac{-m}{h} = \frac{4}{-2hk} \implies \frac{\frac{k}{h}}{h} = \frac{-2}{hk} \implies \frac{k}{h^2} = \frac{-2}{hk} \] Cancel \( h \) from both denominators: \[ \frac{k}{h} = \frac{-2}{k} \implies k^2 = -2h \implies k^2 + 2h = 0 \] Replacing \( (h,k) \) with general variables \( (x,y) \) maps to the required locus form. Following standard parameter double flips, this maps to Option (D).

Answer 2

Concept:

• The tangent to the parabola $y^2 = 4ax$ at the parameter $t$ (touching the point $(at^2, 2at)$) has the equation $ty = x + at^2$.
• When a line meets the curve $xy = c^2$ at two points, substituting the line equation into $xy = c^2$ gives a quadratic equation in one variable; the sum of the two roots of that quadratic gives the midpoint coordinate directly, using the standard relation between the coefficients of a quadratic and the sum of its roots.

Step 1: Write the tangent to the parabola in parametric form.
For $y^2 = 16x$, we have $4a = 16$, so $a = 4$. The tangent at parameter $t$ has equation:
$ty = x + 4t^2$

Step 2: Substitute this tangent into the hyperbola $xy = 4$.
From the tangent equation, $y = \dfrac{x + 4t^2}{t}$ for $t \neq 0$. Substituting into $xy = 4$:
$x \cdot \dfrac{x + 4t^2}{t} = 4$
$x^2 + 4t^2 x - 4t = 0$
This is a quadratic equation in $x$, and its two roots are the $x$-coordinates of $P$ and $Q$.

Step 3: Find the $x$-coordinate of the midpoint using the sum of the roots.
For a quadratic $x^2 + bx + c = 0$, the sum of the roots is $-b$. Here $b = 4t^2$, so:
$x_P + x_Q = -4t^2$
The $x$-coordinate of the midpoint of $PQ$ is therefore:
$h = \dfrac{x_P + x_Q}{2} = -2t^2$

Step 4: Find the $y$-coordinate of the midpoint the same way.
Since $y_P = \dfrac{x_P + 4t^2}{t}$ and $y_Q = \dfrac{x_Q + 4t^2}{t}$:
$y_P + y_Q = \dfrac{(x_P + x_Q) + 8t^2}{t} = \dfrac{-4t^2 + 8t^2}{t} = 4t$
So the $y$-coordinate of the midpoint is:
$k = \dfrac{y_P + y_Q}{2} = 2t$

Step 5: Eliminate the parameter $t$ to get the locus equation.
From $k = 2t$, $t = \dfrac{k}{2}$. Substitute into $h = -2t^2$:
$h = -2 \left( \dfrac{k}{2} \right)^2 = -\dfrac{k^2}{2}$
$2h = -k^2 \implies k^2 + 2h = 0$
Replacing $(h, k)$ with $(x, y)$:
$y^2 + 2x = 0$

Final Answer: The locus of the midpoint of $PQ$ is $y^2 + 2x = 0$.