Question Statement

Part (a): A tangent line is drawn to the parabola at any point $P$ . Prove that the line segment of the tangent cut off between $P$ and the directrix subtends a right angle at the focus.

Part (b): Prove that tangents at the end points of any focal chord intersect at right angles on the directrix.

Background and Explanation

To solve these problems, we use the standard form of a parabola $y^{2} = 4 a x$ and its parametric representation $P (a t^{2}, 2 a t)$ . Key concepts include finding the equation of a tangent using derivatives, the properties of the focus $(a, 0)$ and directrix $x = - a$ , and the condition that two lines are perpendicular if the product of their slopes is $- 1$ .

Solution

Part (a)

Consider the parabola $y^{2} = 4 a x$ and let $P (a t^{2}, 2 a t)$ be any point on the parabola. The directrix of this parabola is: $x = - a or x + a = 0$

To find the slope of the tangent, we differentiate $y^{2} = 4 a x$ with respect to $x$ : $2 y \frac{d y}{d x} = 4 a ⟹ \frac{d y}{d x} = \frac{2 a}{y}$

At the point $P (a t^{2}, 2 a t)$ , the slope ( $m$ ) of the tangent line is: $m = \frac{2 a}{2 a t} = \frac{1}{t}$

The equation of the tangent line at point $P$ is: $y - 2 a t = \frac{1}{t} (x - a t^{2})$ $t y - 2 a t^{2} = x - a t^{2}$ $x - t y + a t^{2} = 0$

Let $Q$ be the point where this tangent intersects the directrix. To find $Q$ , we solve equations (i) and (ii) simultaneously. Substituting $x = - a$ into (ii): $- a - t y + a t^{2} = 0$ $t y = a t^{2} - a$ $y = a (\frac{t ^{2} - 1}{t}) = a t - \frac{a}{t}$

Thus, the point of intersection is $Q (- a, a (t - \frac{1}{t}))$ . The focus of the parabola is $F (a, 0)$ . We now find the slopes of the lines $F Q$ and $F P$ .

Slope of $F Q$ ( $m_{1}$ ): $m_{1} = \frac{( a t - \frac{a}{t} ) - 0}{- a - a} = \frac{a ( t - \frac{1}{t} )}{- 2 a} = - \frac{1}{2} (\frac{t ^{2} - 1}{t})$

Slope of $F P$ ( $m_{2}$ ): $m_{2} = \frac{2 a t - 0}{a t ^{2} - a} = \frac{2 a t}{a ( t ^{2} - 1 )} = \frac{2 t}{t ^{2} - 1}$

Product of Slopes: $m_{1} \cdot m_{2} = (- \frac{t ^{2} - 1}{2 t}) (\frac{2 t}{t ^{2} - 1}) = - 1$

Since the product of the slopes is $- 1$ , $F Q ⊥ F P$ . This proves that the segment $P Q$ subtends a right angle ( $9 0^{\circ}$ ) at the focus.

Part (b)

Consider the parabola $y^{2} = 4 a x$ . Let $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ be the endpoints of a focal chord. For a focal chord, the relationship between the parameters is: $t_{1} t_{2} = - 1$

As derived in Part (a), the slope of the tangent at any point $t$ is $1/ t$ .

Slope of tangent at $P$ ( $m_{P}$ ) = $1/ t_{1}$
Slope of tangent at $Q$ ( $m_{Q}$ ) = $1/ t_{2}$

Product of slopes: $m_{P} \cdot m_{Q} = (\frac{1}{t _{1}}) (\frac{1}{t _{2}}) = \frac{1}{t _{1} t _{2}} = \frac{1}{- 1} = - 1$ This proves the tangents intersect at a right angle.

To find the point of intersection, we use the tangent equations:

$x - t_{1} y + a t_{1}^{2} = 0$
$x - t_{2} y + a t_{2}^{2} = 0$

Subtracting equation (2) from (1): $(t_{2} - t_{1}) y + a (t_{1}^{2} - t_{2}^{2}) = 0$ $(t_{2} - t_{1}) y + a (t_{1} - t_{2}) (t_{1} + t_{2}) = 0$ $y = a (t_{1} + t_{2})$

Substitute $y$ back into equation (1): $x - t_{1} (a (t_{1} + t_{2})) + a t_{1}^{2} = 0$ $x - a t_{1}^{2} - a t_{1} t_{2} + a t_{1}^{2} = 0$ $x = a t_{1} t_{2}$

Since $t_{1} t_{2} = - 1$ , we get: $x = a (- 1) = - a$

The point of intersection is $(- a, a (t_{1} + t_{2}))$ . Since the $x$ -coordinate is $- a$ , this point lies exactly on the directrix. Thus, the tangents intersect at right angles on the directrix.

Key Formulas or Methods Used

Parametric Coordinates: $P (a t^{2}, 2 a t)$ for $y^{2} = 4 a x$ .
Tangent Equation: $t y = x + a t^{2}$ .
Focal Chord Property: $t_{1} t_{2} = - 1$ .
Perpendicularity Condition: $m_{1} \cdot m_{2} = - 1$ .
Directrix Equation: $x = - a$ .
Focus Coordinates: $F (a, 0)$ .

Summary of Steps

Define Points: Use parametric forms for points on the parabola.
Find Tangents: Differentiate the parabola equation to find the slope and then the equation of the tangent line.
Find Intersections: Solve the tangent equation simultaneously with the directrix ( $x = - a$ ) or with another tangent line.
Calculate Slopes: Use the coordinates of the intersection points and the focus to find the slopes of the required line segments.
Verify Orthogonality: Multiply the slopes to show the product is $- 1$ , confirming a $9 0^{\circ}$ angle.