JEE Main 2023 (13 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $P Q$ be a focal chord of the parabola $y^{2} = 36 x$ of length 100, making an acute angle with the positive $x$ -axis. Let the ordinate of $P$ be positive and $M$ be the point on the line segment $P Q$ such that $P M : M Q = 3 : 1$ . Then which of the following points does NOT lie on the line passing through $M$ and perpendicular to the line $P Q$ ?

Q: Why do we get two values for $t$ and how do I know which one to pick?

The quadratic equation $3t^2 - 10t + 3 = 0$ naturally yields two roots, $t=3$ and $t=1/3$. Each corresponds to a different orientation of the chord. We use the slope condition: the chord makes an acute angle with the positive x-axis, meaning the slope must be positive. Calculating the slope $m = \frac{2}{t - 1/t}$ for both values will reveal which one satisfies the condition.

Q: How do I know which point is $P$ and which is $Q$?

The problem specifies the ordinate of $P$ is positive. Since $y = 2at$, $t$ must be positive. Both roots are positive, so we rely on the slope constraint to fix the orientation of the chord, which then determines the coordinates of $P$ and $Q$.

Select Answer:

Visualized Solution

Identify Parabola Parameters

Given Parabola: $y^{2} = 36 x$
Standard Form: $y^{2} = 4 a x$
Comparing coefficients: $4 a = 36 ⟹ a = 9$
Focus $S = (a, 0) = (9, 0)$

Focal Chord Length Formula

Length of focal chord $P Q = L = 100$
Formula for focal chord length: $L = a (t + \frac{1}{t})^{2}$
where $t$ is the parameter of point $P (a t^{2}, 2 a t)$

Setup Equation for $t$

Substitute $a = 9, L = 100$ : $9 (t + \frac{1}{t})^{2} = 100$
Taking square root: $t + \frac{1}{t} = \frac{10}{3}$
Note: $t + \frac{1}{t}$ must be positive since $t$ and $1/ t$ have the same sign.

Solve for Parameter $t$

Rearrange to quadratic form: $3 t^{2} - 10 t + 3 = 0$
Factorize: $3 t^{2} - 9 t - t + 3 = 0$
$(3 t - 1) (t - 3) = 0$
Roots: $t = 3$ or $t = \frac{1}{3}$

Determine Valid Parameter

Condition 1: Ordinate of $P > 0 ⟹ 2 a t > 0 ⟹ t > 0$ (Both valid)
Condition 2: Acute angle with x-axis $⟹$ Slope of $P Q > 0$
Slope of focal chord $m = \frac{2}{t - \frac{1}{t}}$
For $t = 3$ , $m = \frac{2}{3 - \frac{1}{3}} = \frac{3}{4} > 0$ (Acute)
Selected parameter: $t = 3$

Coordinates of $P$ and $Q$

Point $P = (a t^{2}, 2 a t)$
Substitute $a = 9, t = 3$ : $P = (9 \cdot 3^{2}, 2 \cdot 9 \cdot 3) = (81, 54)$
Point $Q = (\frac{a}{t ^{2}}, - \frac{2 a}{t})$
Substitute $a = 9, t = 3$ : $Q = (\frac{9}{3 ^{2}}, - \frac{2 \cdot 9}{3}) = (1, - 6)$

Find Point $M$

Point $M$ divides $P Q$ such that $P M : M Q = 3 : 1$
Section Formula: $M = (\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n})$
Here, $m = 3, n = 1$ , $(x_{1}, y_{1}) = P (81, 54)$ , $(x_{2}, y_{2}) = Q (1, - 6)$

Calculate Coordinates of $M$

$x_{M} = \frac{1 \cdot 81 + 3 \cdot 1}{3 + 1} = \frac{81 + 3}{4} = \frac{84}{4} = 21$
$y_{M} = \frac{1 \cdot 54 + 3 \cdot ( - 6 )}{3 + 1} = \frac{54 - 18}{4} = \frac{36}{4} = 9$
Therefore, $M = (21, 9)$

Slope of Perpendicular Line

Slope of $P Q$ ( $m_{1}$ ): $\frac{54 - ( - 6 )}{81 - 1} = \frac{60}{80} = \frac{3}{4}$
We need a line perpendicular to $P Q$ .
Condition for perpendicularity: $m_{1} \cdot m_{2} = - 1$
Slope of required line ( $m_{2}$ ): $- \frac{1}{\frac{3}{4}} = - \frac{4}{3}$

Equation of the Perpendicular Line

Point-Slope Form: $y - y_{M} = m_{2} (x - x_{M})$
Substitute $M (21, 9)$ and $m_{2} = - \frac{4}{3}$ :
$y - 9 = - \frac{4}{3} (x - 21)$
$3 (y - 9) = - 4 (x - 21)$
$3 y - 27 = - 4 x + 84 ⟹ 4 x + 3 y = 111$

Verify Options

We need to find which point does NOT lie on $4 x + 3 y = 111$ .
Option 1: $(- 6, 45) ⟹ 4 (- 6) + 3 (45) = - 24 + 135 = 111$ (Lies)
Option 2: $(6, 29) ⟹ 4 (6) + 3 (29) = 24 + 87 = 111$ (Lies)
Option 3: $(3, 33) ⟹ 4 (3) + 3 (33) = 12 + 99 = 111$ (Lies)
Option 4: $(- 3, 43) ⟹ 4 (- 3) + 3 (43) = - 12 + 129 = 117 \neq = 111$ (Does NOT lie)

Final Conclusion

The point $(- 3, 43)$ does not lie on the perpendicular line.
Key Concepts Mastered:
1. Focal chord length formula: $a (t + \frac{1}{t})^{2}$
2. Parameter selection based on geometric constraints.
3. Internal section formula and perpendicular line slopes.

00:00 / 00:00

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Geometry of the Focal Chord

Welcome, future engineer! Today, we are embarking on a journey through the elegant world of coordinate geometry. We are looking at a parabola,

y^{2} = 36 x

, and a focal chord

P Q

of length

100

This is a test of your ability to visualize the interplay between algebraic parameters and geometric constraints.

Unlocking the Focal Chord

First, let's identify our parabola. By comparing

y^{2} = 36 x

with the standard form

y^{2} = 4 a x

, we immediately see that

4 a = 36

, which gives us

a = 9

. The focus

S

is at

(9, 0)

The focal chord length formula is a powerful tool in your arsenal:

L = a (t + \frac{1}{t})^{2}

Substituting

a = 9

and

L = 100

, we get:

9 (t + \frac{1}{t})^{2} = 100

This simplifies to:

(t + \frac{1}{t})^{2} = \frac{100}{9} \Rightarrow t + \frac{1}{t} = \frac{10}{3}

The Parameter Dilemma

Solving the quadratic equation

3 t^{2} - 10 t + 3 = 0

gives us two potential parameters:

t = 3

t = \frac{1}{3}

The problem states the chord makes an acute angle with the positive

x

-axis. The slope of the focal chord is given by:

m = \frac{2}{t - \frac{1}{t}}

For

t = 3

, we calculate:

m = \frac{2}{3 - 1/3} = \frac{2}{8/3} = \frac{3}{4}

Since this slope is positive,

t = 3

is our correct parameter.

The Section Formula

With

t = 3

and

a = 9

, we find the coordinates of the endpoints:

P = (a t^{2}, 2 a t) = (81, 54)

Q = (\frac{a}{t ^{2}}, - \frac{2 a}{t}) = (1, - 6)

Now, we need point

M

P Q

such that

P M : M Q = 3 : 1

. Using the internal section formula:

M = (\frac{1 \cdot 81 + 3 \cdot 1}{4}, \frac{1 \cdot 54 + 3 \cdot ( - 6 )}{4}) = (21, 9)

The Perpendicular Line

Finally, we need the line through

M (21, 9)

perpendicular to

P Q

. Since the slope of

P Q

3/4

, the slope of our required line is

- 4/3

The equation is:

y - 9 = - \frac{4}{3} (x - 21)

Multiplying through and rearranging, we obtain:

3 y - 27 = - 4 x + 84

4 x + 3 y = 111

Testing the point

(- 3, 43)

in the equation

4 x + 3 y

4 (- 3) + 3 (43) = - 12 + 129 = 117

The final equation of the line is $4 x + 3 y = 111$ . Keep practicing, and you will master these concepts.

Common Traps & FAQs

Why do we get two values for

t

and how do I know which one to pick?

The quadratic equation

3 t^{2} - 10 t + 3 = 0

naturally yields two roots,

t = 3

and

t = 1/3

. Each corresponds to a different orientation of the chord. We use the slope condition: the chord makes an acute angle with the positive x-axis, meaning the slope must be positive. Calculating the slope

m = \frac{2}{t - 1/ t}

for both values will reveal which one satisfies the condition.

How do I know which point is

P

and which is

Q

The problem specifies the ordinate of

P

is positive. Since

y = 2 a t

t

must be positive. Both roots are positive, so we rely on the slope constraint to fix the orientation of the chord, which then determines the coordinates of

P

and

Q

Is there a faster way to find the perpendicular line?

While there are properties related to the normal of a parabola, the standard coordinate geometry approach—finding the slope of

P Q

, determining the perpendicular slope, and using the point-slope form—is robust and less prone to conceptual errors. Stick to the fundamentals!

Select Answer:

Visualized Solution

Identify Parabola Parameters

Focal Chord Length Formula

Setup Equation for $t$

Solve for Parameter $t$

Determine Valid Parameter

Coordinates of $P$ and $Q$

Find Point $M$

Calculate Coordinates of $M$

Slope of Perpendicular Line

Equation of the Perpendicular Line

Verify Options

Final Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Geometry of the Focal Chord

Unlocking the Focal Chord

The Parameter Dilemma

The Section Formula

The Perpendicular Line

Similar Questions

Let $R$ be the focus of the parabola $y^{2} = 20 x$ and the line $y = m x + c$ intersect the parabola at two points $P$ and $Q$ . Let the points $G (10, 10)$ be the centroid of the triangle $P QR$ . If $c - m = 6$ , then $P Q^{2}$ is

Let the length of the focal chord $P Q$ of the parabola $y^{2} = 12 x$ be 15 units. If the distance of $P Q$ from the origin is $p$ , then $10 p^{2}$ is equal to _______

Let $P$ and $Q$ be distinct points on the parabola $y^{2} = 2 x$ such that a circle with $P Q$ as diameter passes through the vertex $O$ of the parabola. If $P$ lies in the first quadrant and the area of the triangle $Δ O P Q$ is $32$ , then which of the following is (are) the coordinates of $P$ ?

Let $P (4, 43)$ be a point on the parabola $y^{2} = 4 a x$ and PQ be a focal chord of the parabola. If M and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:

Let $P$ be a variable point on the parabola $y = 4 x^{2} + 1$ . Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is :

Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin on the positive x-axis then which of the following points does not lie on it ?

Let the point $P$ of the focal chord $P Q$ of the parabola $y^{2} = 16 x$ be $(1, - 4)$ . If the focus of the parabola divides the chord $P Q$ in the ratio $m : n$ , $gcd (m, n) = 1$ , then $m^{2} + n^{2}$ is equal to :

Let $O$ be the vertex of the parabola $x^{2} = 4 y$ and $Q$ be any point on it. Let the locus of the point $P$ , which divides the line segment $O Q$ internally in the ratio $2 : 3$ be the conic $C$ . Then the equation of the chord of $C$ , which is bisected at the point $(1, 2)$ , is :

Select Answer:

Visualized Solution

Identify Parabola Parameters

Focal Chord Length Formula

Setup Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Solve for Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Determine Valid Parameter

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Find Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Calculate Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Slope of Perpendicular Line

Equation of the Perpendicular Line

Verify Options

Final Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Geometry of the Focal Chord

Unlocking the Focal Chord

The Parameter Dilemma

The Section Formula

The Perpendicular Line

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the length of the focal chord PQ of the parabola y2=12x be 15 units. If the distance of PQ from the origin is p, then 10p2 is equal to _______

Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin on the positive x-axis then which of the following points does not lie on it ?

Setup Equation for $t$

Solve for Parameter $t$

Coordinates of $P$ and $Q$

Find Point $M$

Calculate Coordinates of $M$

Let the length of the focal chord $P Q$ of the parabola $y^{2} = 12 x$ be 15 units. If the distance of $P Q$ from the origin is $p$ , then $10 p^{2}$ is equal to _______