From $P$ on $b^4x+2a^2y^2=0$ are drawn tangents $PQ$ and $PR$ to $x^2/a^2-y^2/b^2=1$. If $QR$ meets a fixed parabola, find the parabola's equation.

Question

A JEE advanced level question:

From a point $P$ on the curve $b^4x + 2a^2y^2=0$, a pair of tangents $PQ$ and $PR$ are drawn to hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} =1$. ($a$, $b$ are some constants.) If $QR$ touches a fixed parabola, then the equation of the parabola is ...

Options are: (i)$\;x^2=4y \qquad$(ii)$\;y^2=8x\qquad$(iii)$\;y^2=4x\qquad$(iv)$\;x^2=8y$

The correct answer is $y^2=8x$.

Please provide a conceptual and clear solution.

I have tried to write the given curve as a parabola, with its focus at $(\frac{-b^4}{2a^2},0)$ and a parametric point on that parabola as ($-at^2,2at$) and write pair of tangents from this parabola to a hyperbola, but it became too complicated as it is very long to solve the equation of the tangents with the given hyperbola, and even if I get to know the points where the tangent touches the hyperbola, how do I know which parabola it always touches, please provide a better conceptual approach.

Answer 1

(Geogebra figure) We have taken here $a=3$ and $b=4$. Parabola (P) with equ. (1) or (2) in red, Hyperbola (H) with equation (3a) in blue. Envelope of lines $QR$ is (P') in green.

Parabola (P) has cartesian equation :

$$b^4x+2a^2y^2=0\tag{1}$$

Let us work on it with (equivalent !) parametric equations :

$$(x=-2a^2t^2,y=b^2t)\tag{2}$$

Line $QR$ connecting the two points of tangency issued from a generic point $P(x_0,y_0)$ on parabola $(P)$ is called the polar line of point $P$ wrt to hyperbola $(H)$ ; the equation of this polar line can be obtained directly (see (3) below) : you haven't to bother about 1) computing the equations of the tangent lines (which could have been obtained by the $SS_0=T^2$ formula, but that's another story) and 2) computing the coordinates of the points of tangency.

The equation of this polar line is classicaly obtained by taking the bilinear form associated with the equ. of (H):

$$\begin{cases}&x^2/a^2−y^2/b^2−1&=&0 \ & (3a) \\ \longrightarrow & xx_0/a^2−yy_0/b^2−1&=&0 \ & (3b) \end{cases}\tag{3}$$

Using parametric equations (1) into (3b) for $x=x_0,y=y_0$

$$y=-2xt-\frac{1}{t}$$

When $t$ varies, this line envelopes a curve. Let us show that this curve has equation $y^2=8x.$

In general, the envelope of a family of straight lines depending upon a parameter $t$ is obtained using a classical technique involving derivation wrt parameter $t$. (See for example my answer here).

Here we have the system $$\begin{cases}\text{initial equation } \ &y&=&-2xt-\frac{1}{t}\\\text{deriv. wrt parameter t } \ &0&=&-2x+\frac{1}{t^2}\end{cases}.$$ Eliminating $t$, we get

$$x=\frac{1}{t^2},y=-\frac{2}{t}$$ which are parametric equations of parabola

$$x=8y^2\tag{4}$$

giving a direct way to get the final result.

Important remark : In fact, all this question deals with a classical curve transformation which was well studied, say between 1850 and 1950, under the name "(reciprocal) polar transform". See for example here. Here the polar transform (wrt to hyperbola (H)) of (1) is (4).

Answer 2

As the OP mentioned, the parametric equation of point $P$ is

$ P = ( K t^2, t ) $

where $ K = - \dfrac{2 a^2}{b^4} $

Now the gradient of the hyperbola at any point $(x, y)$ on it is given by

$ g = 2 ( \dfrac{x}{a^2} , - \dfrac{y}{b^2} ) $

Since we want the tangent line to pass through $P$, then

$ g \cdot ( x - K t^2 , y - t ) = 0 $

where $\cdot$ is the dot product. Hence,

$ \dfrac{x}{a^2} (x - K t^2) - \dfrac{y}{b^2} (y - t ) = 0 $

Using $ \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 $ the above reduces to

$ 1 - \dfrac{K x t^2}{a^2} + \dfrac{y t}{b^2} = 0 $

And this is an equation of the line on which points $Q$ and $R$ lie. We don't need to find $Q$ and $R$ explicitly. If we did, then all we have to do is intersect the equation of the line with the equation of the hyperbola. However, as I said, we don't need that. All we need is the equation of the line that passes through $Q$ and $R$.

Now let

$ f(x, y, t) = 1 - \dfrac{K x t^2}{a^2} + \dfrac{y t}{b^2} = 0 $

We want to find the envelope of all these lines as $t$ changes. To do this, we find the partial derivative of $f$ with respect to $t$ and set that equal to $0$ in order to solve for $t$ as a function of $x$ and $y$.

$\dfrac{\partial f}{\partial t} = - \dfrac{2 K x t }{a^2} + \dfrac{y}{b^2} = 0 $

Then, solving for $t$ we get

$ t = \dfrac{ a^2 y }{ 2 b^2 K x } $

Recall that $ K = -\dfrac{2 a^2}{b^4} $, therefore,

$ t = - \dfrac{ b^2 y } { 4 x } $

Plug this into the expression of $ f(x, y, t) = 0 $, you get

$1 - \left( \dfrac{K x }{a^2} \right) \left( \dfrac{b^4 y^2}{16 x^2} \right) + \left(\dfrac{y}{b^2} \right) \left( - \dfrac{ b^2 y }{4 x} \right) = 0 $

After plugging in the expression for $K$, this reduces to

$ 1 + \dfrac{ y^2 }{8 x} - \dfrac{ y^2 }{4 x} = 0 $

And this reduces further to,

$ 1 - \dfrac{ y^2}{8 x} = 0 $

which simplifies to

$ y^2 = 8 x $

And this the equation of the envelope of all lines $QR$.