Question Statement

Let $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ be any two points on the parabola $y^{2} = 4 a x$ . Prove that the chord joining $P$ and $Q$ is a focal chord if $t_{1} t_{2} = - 1$ .

Background and Explanation

A focal chord is a line segment that passes through the focus of a conic section. For the standard parabola $y^{2} = 4 a x$ , the focus is located at the point $(a, 0)$ . To prove the condition, we find the equation of the line passing through two points and ensure it satisfies the coordinates of the focus.

Solution

To prove the given condition, we first determine the equation of the line (chord) passing through the points $P$ and $Q$ , and then apply the condition that it must pass through the focus $(a, 0)$ .

Step 1: Find the equation of the chord $P Q$

The points are given as $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ . Using the two-point formula for a straight line:

$\frac{y - y _{1}}{y _{2} - y _{1}} = \frac{x - x _{1}}{x _{2} - x _{1}}$

Substituting the coordinates of $P$ and $Q$ :

$\frac{x - a t _{1}^{2}}{a t _{2}^{2} - a t _{1}^{2}} (2 a t_{2} - 2 a t_{1}) (x - a t_{1}^{2}) = \frac{y - 2 a t _{1}}{2 a t _{2} - 2 a t _{1}} = (a t_{2}^{2} - a t_{1}^{2}) (y - 2 a t_{1})$

Now, we simplify the equation by factoring out common terms:

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

Dividing both sides by $a (t_{2} - t_{1})$ (assuming $t_{1} \neq = t_{2}$ ):

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

Step 2: Apply the focal chord condition

A chord is a focal chord if it passes through the focus $(a, 0)$ . We substitute $x = a$ and $y = 0$ into the simplified equation of the chord:

$2 (a - a t_{1}^{2}) 2 a (1 - t_{1}^{2}) = (t_{2} + t_{1}) (0 - 2 a t_{1}) = (t_{2} + t_{1}) (- 2 a t_{1})$

Divide both sides by $2 a$ :

$1 - t_{1}^{2} 1 - t_{1}^{2} = - t_{1} (t_{2} + t_{1}) = - t_{1} t_{2} - t_{1}^{2}$

Adding $t_{1}^{2}$ to both sides:

$1 = - t_{1} t_{2}$

Which simplifies to:

$t_{1} t_{2} = - 1$

Hence proved.

Key Formulas or Methods Used

Two-point formula for a line: $\frac{y - y _{1}}{x - x _{1}} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Parametric coordinates of a parabola: $(a t^{2}, 2 a t)$
Focus of $y^{2} = 4 a x$ : $(a, 0)$
Algebraic Identity: $a^{2} - b^{2} = (a - b) (a + b)$

Summary of Steps

Set up the equation of the chord $P Q$ using the two-point form with parametric coordinates.
Simplify the equation by factoring and canceling common terms like $a (t_{2} - t_{1})$ .
Substitute the coordinates of the focus $(a, 0)$ into the line equation, as a focal chord must pass through this point.
Perform algebraic simplification to isolate the product $t_{1} t_{2}$ , resulting in $t_{1} t_{2} = - 1$ .

Question Statement

Background and Explanation

Solution

To prove the given condition, we first determine the equation of the line (chord) passing through the points $P$ and $Q$ , and then apply the condition that it must pass through the focus $(a, 0)$ .

Step 1: Find the equation of the chord $P Q$

The points are given as $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ . Using the two-point formula for a straight line:

$\frac{y - y _{1}}{y _{2} - y _{1}} = \frac{x - x _{1}}{x _{2} - x _{1}}$

Substituting the coordinates of $P$ and $Q$ :

$\frac{x - a t _{1}^{2}}{a t _{2}^{2} - a t _{1}^{2}} (2 a t_{2} - 2 a t_{1}) (x - a t_{1}^{2}) = \frac{y - 2 a t _{1}}{2 a t _{2} - 2 a t _{1}} = (a t_{2}^{2} - a t_{1}^{2}) (y - 2 a t_{1})$

Now, we simplify the equation by factoring out common terms:

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

Dividing both sides by $a (t_{2} - t_{1})$ (assuming $t_{1} \neq = t_{2}$ ):

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

Step 2: Apply the focal chord condition

A chord is a focal chord if it passes through the focus $(a, 0)$ . We substitute $x = a$ and $y = 0$ into the simplified equation of the chord:

$2 (a - a t_{1}^{2}) 2 a (1 - t_{1}^{2}) = (t_{2} + t_{1}) (0 - 2 a t_{1}) = (t_{2} + t_{1}) (- 2 a t_{1})$

Divide both sides by $2 a$ :

$1 - t_{1}^{2} 1 - t_{1}^{2} = - t_{1} (t_{2} + t_{1}) = - t_{1} t_{2} - t_{1}^{2}$

Adding $t_{1}^{2}$ to both sides:

$1 = - t_{1} t_{2}$

Which simplifies to:

$t_{1} t_{2} = - 1$

Hence proved.

Key Formulas or Methods Used

Two-point formula for a line: $\frac{y - y _{1}}{x - x _{1}} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Parametric coordinates of a parabola: $(a t^{2}, 2 a t)$
Focus of $y^{2} = 4 a x$ : $(a, 0)$
Algebraic Identity: $a^{2} - b^{2} = (a - b) (a + b)$

Summary of Steps

Set up the equation of the chord $P Q$ using the two-point form with parametric coordinates.
Simplify the equation by factoring and canceling common terms like $a (t_{2} - t_{1})$ .
Substitute the coordinates of the focus $(a, 0)$ into the line equation, as a focal chord must pass through this point.
Perform algebraic simplification to isolate the product $t_{1} t_{2}$ , resulting in $t_{1} t_{2} = - 1$ .

7.5 Q-8

Question Statement

Background and Explanation

Solution

Step 1: Find the equation of the chord $P Q$

Step 2: Apply the focal chord condition

Key Formulas or Methods Used

Summary of Steps

7.5 Q-8

Question Statement

Background and Explanation

Solution

Step 1: Find the equation of the chord $P Q$

Step 2: Apply the focal chord condition

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Step 1: Find the equation of the chord PQ

Step 2: Apply the focal chord condition

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Step 1: Find the equation of the chord PQ

Step 2: Apply the focal chord condition

Key Formulas or Methods Used

Summary of Steps

Step 1: Find the equation of the chord $P Q$

Step 1: Find the equation of the chord $P Q$