Animated Solution for Mathematics - Conic Sections: If a chord, which is not a tangent, of the parabola $y^2=16x$ has the equation $2x+y=p$, and midpoint $(h,k)$, then which of the following is(are) possible value(s) of $p,h$ and $k$?

Imagine you are standing before the parabola $y^2=16x$, a classic curve opening gracefully to the right. A chord slices through it, defined by the linear equation $2x+y=p$.

We are told this chord has a midpoint at $(h,k)$. Our mission is to uncover the possible values for $p,h,$ and $k$. This is not just an algebra problem; it is a beautiful exercise in the symmetry of conic sections.

In the world of JEE Advanced, when you encounter a chord with a known midpoint, there is one tool that stands above all others: the $T=S_1$ formula. This is the golden key.

$T$ represents the equation of the tangent at the point $(h,k)$, and $S_1$ is the value of the parabola's equation at that same point. To find $T$, we take the equation of our parabola, $y^2-16x=0$, and perform the standard transformations: replace $y^2$ with $yk$ and $x$ with $\frac{x+h}{2}$.

This yields:

This simplifies beautifully to:

Next, we find $S_1$ by simply substituting the midpoint $(h,k)$ into the parabola's equation:

Now, we equate $T$ and $S_1$:

Expanding the left side, we get $yk-8x-8h=k^2-16h$. Rearranging this into the standard form of a straight line, $Ax+By=C$, we move the $x$ and $y$ terms to the left and the constants to the right:

We now have two equations for the same chord: the one given in the problem, $2x+y=p$, and our derived equation, $-8x+ky=k^2-8h$. Because these lines are identical, their coefficients must be proportional.

This leads us to the elegant ratio:

This is where the magic happens. Looking at the first two parts of our proportion, $\frac{-8}{2}=\frac{k}{1}$, we immediately see that:

With $k$ in hand, we turn to the second part of the proportion:

Substituting $k=-4$, we get:

This simplifies to:

Multiplying by $p$ and dividing by $-4$, we arrive at the linear relationship:

By testing our options against these two conditions—$k=-4$ and $2h-p=4$—we find that only option D satisfies both. It is a perfect, logical conclusion to a beautiful geometric journey.