JEE Advanced 2013

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Comprehension Passage

Q: How do we know the product $t_1 t_2 = -1$?

A focal chord passes through the focus $(a, 0)$. The line passing through $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ has a slope $m = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2}{t_1+t_2}$. The equation of the line is $y - 2at_1 = \frac{2}{t_1+t_2}(x - at_1^2)$. Since it passes through $(a, 0)$, we have $-2at_1 = \frac{2}{t_1+t_2}(a - at_1^2)$. Simplifying this leads directly to $t_1t_2 = -1$.

Q: Why is the angle $\tan \theta$ negative in the final answer?

The angle $\theta$ is the angle between the lines $VP$ and $VQ$. The formula $\tan \theta = \frac{m_1 - m_2}{1 + m_1m_2}$ gives the tangent of the angle between two lines. Depending on the order of $m_1$ and $m_2$, the result can be positive or negative. Since the question asks for the value and provides options, we choose the one that matches our calculation based on the geometry of the chord.

Let

P Q

be a focal chord of the parabola

y^{2} = 4 a x

. The tangents to the parabola at

P

and

Q

meet at a point lying on the line

y = 2 x + a, a > 0

Question 1:

Length of chord $P Q$ is

Select Answer:

Question 2:

If chord $P Q$ subtends an angle $θ$ at the vertex of $y^{2} = 4 a x$ , then $tan θ =$

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the parabola

y^{2} = 4 a x

and its focus

F (a, 0)

. A focal chord

P Q

is a line passing through

F

. The key geometric insight is that the tangents at the endpoints of any focal chord always intersect on the directrix of the parabola. Can you identify the equation of the directrix?

Hint 2

The Arsenal: Use the parametric coordinates

P (a t_{1}^{2}, 2 a t_{1})

and

Q (a t_{2}^{2}, 2 a t_{2})

. Recall the fundamental property of focal chords: the product of the parameters is

t_{1} t_{2} = - 1

. This is your most powerful tool to simplify the intersection point of the tangents.

Hint 3

The Execution: The intersection point

R

of the tangents at

P

and

Q

is given by

R (a t_{1} t_{2}, a (t_{1} + t_{2}))

. Since

t_{1} t_{2} = - 1

, the coordinates of

R

become

(- a, a (t_{1} + t_{2}))

. Substitute these coordinates into the given line equation

y = 2 x + a

to find the value of

t_{1} + t_{2}

Visualized Solution

Visualizing the Geometry

Parabola: $y^{2} = 4 a x$
Focus: $F (a, 0)$
Focal Chord: $P Q$ passes through $F$

Parametric Coordinates

Let $P = (a t_{1}^{2}, 2 a t_{1})$
Let $Q = (a t_{2}^{2}, 2 a t_{2})$
Focal Chord Property: $t_{1} t_{2} = - 1$

Intersection of Tangents

Tangents at $P$ and $Q$ intersect at $R$
Standard Formula: $R = (a t_{1} t_{2}, a (t_{1} + t_{2}))$
Substitute $t_{1} t_{2} = - 1$ : $R = (- a, a (t_{1} + t_{2}))$

Applying the Line Constraint

Point $R (- a, a (t_{1} + t_{2}))$ lies on the given line
Line Equation: $y = 2 x + a$
Substitution: $a (t_{1} + t_{2}) = 2 (- a) + a$

Solving for the Sum of Parameters

$a (t_{1} + t_{2}) = - 2 a + a$
$a (t_{1} + t_{2}) = - a$
Result: $t_{1} + t_{2} = - 1$

Length of Focal Chord Formula

Formula: Length $P Q = a (t_{1} - t_{2})^{2}$
Algebraic Identity: $(t_{1} - t_{2})^{2} = (t_{1} + t_{2})^{2} - 4 t_{1} t_{2}$

Calculating the Length

Substitute $t_{1} + t_{2} = - 1$ and $t_{1} t_{2} = - 1$
$(t_{1} - t_{2})^{2} = (- 1)^{2} - 4 (- 1)$
$(t_{1} - t_{2})^{2} = 1 + 4 = 5$
Final Length: $P Q = 5 a$

Angle at the Vertex

Vertex $V = (0, 0)$
Connect $V$ to $P$ and $Q$
Let $θ$ be the angle subtended by $P Q$ at $V$

Slopes of $V P$ and $V Q$

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

The $tan θ$ Formula

Angle Formula: $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$
Substitute Slopes: $tan θ = \frac{\frac{2}{t _{1}} - \frac{2}{t _{2}}}{1 + ( \frac{2}{t _{1}} ) ( \frac{2}{t _{2}} )}$

Simplifying the Expression

Numerator: $\frac{2 ( t _{2} - t _{1} )}{t _{1} t _{2}}$
Denominator: $1 + \frac{4}{t _{1} t _{2}} = \frac{t _{1} t _{2} + 4}{t _{1} t _{2}}$
Simplified: $tan θ = \frac{2 ( t _{2} - t _{1} )}{t _{1} t _{2} + 4}$

Final Calculation for $tan θ$

We know $t_{1} t_{2} = - 1$
We found $(t_{1} - t_{2})^{2} = 5 ⟹ t_{2} - t_{1} = \pm 5$
From options, taking negative root: $t_{2} - t_{1} = - 5$
$tan θ = \frac{2 ( - 5 )}{- 1 + 4} = - \frac{2}{3} 5$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The parabola is defined by the equation

y^{2} = 4 a x

. The focus of this parabola is located at the point

F (a, 0)

We consider a focal chord

P Q

passing through the focus

F

. The points

P

and

Q

are represented by the parametric coordinates

P (a t_{1}^{2}, 2 a t_{1})

and

Q (a t_{2}^{2}, 2 a t_{2})

For any chord passing through the focus, the parameters satisfy the fundamental constraint:

t_{1} t_{2} = - 1

The Intersection of Tangents

The tangents at points

P

and

Q

intersect at a point

R

. The coordinates of this intersection point are given by:

R (a t_{1} t_{2}, a (t_{1} + t_{2}))

Substituting the focal chord constraint

t_{1} t_{2} = - 1

into the coordinates of

R

, we find the

x

-coordinate to be

- a

. This confirms that the intersection point

R

always lies on the directrix,

x = - a

Solving the Constraint

We are given that the point

R

lies on the line

y = 2 x + a

. Substituting the coordinates of

R (- a, a (t_{1} + t_{2}))

into this line equation, we obtain:

a (t_{1} + t_{2}) = 2 (- a) + a

Simplifying the right side, we get

a (t_{1} + t_{2}) = - a

. Dividing by

a

(assuming $a eq 0$), we arrive at the result:

t_{1} + t_{2} = - 1

The Length of the Chord

The length of a focal chord

P Q

is given by the formula:

L = a (t_{1} - t_{2})^{2}

Using the algebraic identity

(t_{1} - t_{2})^{2} = (t_{1} + t_{2})^{2} - 4 t_{1} t_{2}

, we substitute our known values:

(t_{1} - t_{2})^{2} = (- 1)^{2} - 4 (- 1) = 1 + 4 = 5

Thus, the length of the focal chord

P Q

is $5 a$ .

The Angle at the Vertex

The slopes of the lines

V P

and

V Q

(where

V

is the vertex

(0, 0)

) are

m_{1} = \frac{2}{t _{1}}

and

m_{2} = \frac{2}{t _{2}}

. The angle

θ

subtended at the vertex is determined by:

tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}

Substituting the slopes, we get:

tan θ = \frac{\frac{2}{t _{1}} - \frac{2}{t _{2}}}{1 + \frac{4}{t _{1} t _{2}}} = \frac{2 ( t _{2} - t _{1} )}{t _{1} t _{2} + 4}

Using

t_{1} t_{2} = - 1

and the difference

(t_{2} - t_{1}) = \pm (t_{1} + t_{2})^{2} - 4 t_{1} t_{2} = \pm 5

, we calculate:

tan θ = \frac{2 ( \pm 5 )}{- 1 + 4} = \pm \frac{2 5}{3}

The final result for the tangent of the angle is $\pm \frac{2}{3} 5$ .

Common Traps & FAQs

Why does the intersection of tangents at the ends of a focal chord always lie on the directrix?

This is a beautiful property of parabolas! If you derive the intersection point of tangents at

P (a t_{1}^{2}, 2 a t_{1})

and

Q (a t_{2}^{2}, 2 a t_{2})

, you get

R (a t_{1} t_{2}, a (t_{1} + t_{2}))

. For a focal chord,

t_{1} t_{2} = - 1

, so the x-coordinate of

R

a (- 1) = - a

. Since the directrix of

y^{2} = 4 a x

x = - a

, the point

R

must lie on it. It is not a coincidence; it is the fundamental geometry of the parabola!

How do we know the product

t_{1} t_{2} = - 1

A focal chord passes through the focus

(a, 0)

. The line passing through

P (a t_{1}^{2}, 2 a t_{1})

and

Q (a t_{2}^{2}, 2 a t_{2})

has a slope

m = \frac{2 a t _{2} - 2 a t _{1}}{a t _{2}^{2} - a t _{1}^{2}} = \frac{2}{t _{1} + t _{2}}

. The equation of the line is

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

. Since it passes through

(a, 0)

, we have

- 2 a t_{1} = \frac{2}{t _{1} + t _{2}} (a - a t_{1}^{2})

. Simplifying this leads directly to

t_{1} t_{2} = - 1

Why is the angle

tan θ

negative in the final answer?

The angle

θ

is the angle between the lines

V P

and

V Q

. The formula

tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}

gives the tangent of the angle between two lines. Depending on the order of

m_{1}

and

m_{2}

, the result can be positive or negative. Since the question asks for the value and provides options, we choose the one that matches our calculation based on the geometry of the chord.

Animated Solution for Mathematics - Conic Sections: Comprehension Passage

Length of chord $P Q$ is

Select Answer:

If chord $P Q$ subtends an angle $θ$ at the vertex of $y^{2} = 4 a x$ , then $tan θ =$

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Coordinates

Intersection of Tangents

Applying the Line Constraint

Solving for the Sum of Parameters

Length of Focal Chord Formula

Calculating the Length

Angle at the Vertex

Slopes of $V P$ and $V Q$

The $tan θ$ Formula

Simplifying the Expression

Final Calculation for $tan θ$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Intersection of Tangents

Solving the Constraint

The Length of the Chord

The Angle at the Vertex

Similar Questions

Comprehension Passage

The value of $r$ is

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Animated Solution for Mathematics - Conic Sections: Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Length of chord PQ is

Select Answer:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If chord PQ subtends an angle θ at the vertex of y2=4ax, then tanθ=

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Coordinates

Intersection of Tangents

Applying the Line Constraint

Solving for the Sum of Parameters

Length of Focal Chord Formula

Calculating the Length

Angle at the Vertex

Slopes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VP and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VQ

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ Formula

Simplifying the Expression

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Intersection of Tangents

Solving the Constraint

The Length of the Chord

The Angle at the Vertex

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of r is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If st=1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a line along a chord of the circle 4x2+4y2+120x+675=0, passes through the point (−30,0) and is tangent to the parabola y2=30x, then the length of this chord is :

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Length of chord $P Q$ is

If chord $P Q$ subtends an angle $θ$ at the vertex of $y^{2} = 4 a x$ , then $tan θ =$

Slopes of $V P$ and $V Q$

The $tan θ$ Formula

Final Calculation for $tan θ$

The value of $r$ is

If $s t = 1$ , then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is

If a line along a chord of the circle $4 x^{2} + 4 y^{2} + 120 x + 675 = 0$ , passes through the point $(- 30, 0)$ and is tangent to the parabola $y^{2} = 30 x$ , then the length of this chord is :

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.