JEE Main 2020 (3 September Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $P$ be a point on the parabola, $y^{2} = 12 x$ and $N$ be the foot of the perpendicular drawn from $P$ on the axis of the parabola. A line is now drawn through the mid-point $M$ of $P N$ , parallel to its axis which meets the parabola at $Q$ . If the $y$ -intercept of the line $N Q$ is $\frac{4}{3}$ , then :

Q: Why do we use parametric coordinates $(3t^2, 6t)$ instead of $(x, y)$?

Great question! Using $(x, y)$ introduces two variables, which makes the algebra cumbersome. By using the parametric form $P(at^2, 2at)$, we reduce the entire problem to a single variable $t$. This is a standard, powerful technique in coordinate geometry that simplifies complex intersections into manageable algebraic expressions.

Q: Why is the line through $M$ parallel to the axis simply $y = 3t$?

The axis of the parabola $y^2 = 12x$ is the $x$-axis. A line parallel to the $x$-axis is horizontal, meaning every point on that line has the same $y$-coordinate. Since the line passes through $M(3t^2, 3t)$, its $y$-coordinate must be $3t$ everywhere. Hence, the equation is $y = 3t$.

Select Answer:

Visualized Solution

Identify the Parabola

Given parabola: $y^{2} = 12 x$
Standard form: $y^{2} = 4 a x$
Comparing coefficients: $4 a = 12 ⟹ a = 3$

Define Point $P$ Parametrically

Parametric coordinates of any point on $y^{2} = 4 a x$ are $(a t^{2}, 2 a t)$ .
Substitute $a = 3$ into the coordinates.
Point $P$ becomes $(3 t^{2}, 6 t)$ .

Find Foot of Perpendicular $N$

$N$ is the foot of the perpendicular from $P$ to the $x$ -axis.
The $x$ -coordinate remains the same: $3 t^{2}$ .
The $y$ -coordinate becomes $0$ .
Coordinates of $N$ : $(3 t^{2}, 0)$ .

Locate Midpoint $M$

$M$ is the midpoint of the line segment $P N$ .
Use the midpoint formula: $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ .
$M = (\frac{3 t ^{2} + 3 t ^{2}}{2}, \frac{6 t + 0}{2})$
$M = (3 t^{2}, 3 t)$

Line Through $M$ Parallel to Axis

A line is drawn through $M$ parallel to the $x$ -axis.
A horizontal line has a constant $y$ -coordinate.
Equation of this line: $y = 3 t$ .

Find Coordinates of $Q$

This line meets the parabola $y^{2} = 12 x$ at point $Q$ .
Substitute $y = 3 t$ : $(3 t)^{2} = 12 x$
$9 t^{2} = 12 x ⟹ x = \frac{9 t ^{2}}{12} = \frac{3 t ^{2}}{4}$
Coordinates of $Q$ : $(\frac{3 t ^{2}}{4}, 3 t)$ .

Equation of Line $N Q$

Find the equation of the line passing through $N (3 t^{2}, 0)$ and $Q (\frac{3 t ^{2}}{4}, 3 t)$ .
Slope $m = \frac{3 t - 0}{\frac{3 t ^{2}}{4} - 3 t ^{2}} = \frac{3 t}{- \frac{9 t ^{2}}{4}}$
$m = - \frac{12 t}{9 t ^{2}} = - \frac{4}{3 t}$
Equation of $N Q$ : $y - 0 = - \frac{4}{3 t} (x - 3 t^{2})$

Calculate $y$ -intercept

To find the $y$ -intercept, set $x = 0$ in the equation of $N Q$ .
$y = - \frac{4}{3 t} (0 - 3 t^{2})$
$y = - \frac{4}{3 t} (- 3 t^{2})$
$y$ -intercept $= 4 t$

Solve for Parameter $t$

The problem states that the $y$ -intercept is $\frac{4}{3}$ .
Equate our result to the given value: $4 t = \frac{4}{3}$ .
Solving for $t$ : $t = \frac{1}{3}$ .

Calculate Length $M Q$

Calculate the length of the horizontal segment $M Q$ .
$M Q = ∣ x_{M} - x_{Q} ∣$
$M Q = ∣3 t^{2} - \frac{3 t ^{2}}{4} ∣ = \frac{9 t ^{2}}{4}$

Conclusion and Verification

Substitute $t = \frac{1}{3}$ into the length formula.
$M Q = \frac{9 ( \frac{1}{3} ) ^{2}}{4}$
$M Q = \frac{9 ( \frac{1}{9} )}{4} = \frac{1}{4}$
The correct option is (A).

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

Solution Diagram

Analyzing the Setup

The parabola is defined by the equation

y^{2} = 12 x

. By comparing this to the standard form

y^{2} = 4 a x

, we identify

4 a = 12

, which yields the focal parameter $a = 3$ .

To simplify our calculations, we utilize parametric coordinates. Any point

P

on this parabola can be represented as:

P = (3 t^{2}, 6 t)

This transformation allows us to reduce the complexity of the problem by expressing all geometric features in terms of the single parameter

t

Constructing the Geometry

We drop a perpendicular from

P

to the axis of the parabola (the

x

-axis). The foot of this perpendicular,

N

, shares the same

x

-coordinate as

P

but has a

y

-coordinate of

0

N = (3 t^{2}, 0)

Next, we define

M

as the midpoint of the segment

P N

. Using the midpoint formula, we calculate:

M = (\frac{3 t ^{2} + 3 t ^{2}}{2}, \frac{6 t + 0}{2}) = (3 t^{2}, 3 t)

The Intersection of Lines

A line is drawn through

M

parallel to the

x

-axis. Since this line is horizontal and passes through

M (3 t^{2}, 3 t)

, its equation is simply:

y = 3 t

This line intersects the parabola at point

Q

. Substituting

y = 3 t

into the parabola's equation

y^{2} = 12 x

, we get:

(3 t)^{2} = 12 x \Rightarrow 9 t^{2} = 12 x \Rightarrow x = \frac{3 t ^{2}}{4}

Thus, the coordinates of the intersection point are

Q = (\frac{3 t ^{2}}{4}, 3 t)

The Line $N Q$ and the Climax

We now determine the equation of the line passing through

N (3 t^{2}, 0)

and

Q (\frac{3 t ^{2}}{4}, 3 t)

. The slope

m

of this line is:

m = \frac{3 t - 0}{\frac{3 t ^{2}}{4} - 3 t ^{2}} = \frac{3 t}{- \frac{9 t ^{2}}{4}} = - \frac{4}{3 t}

Using the point-slope form

y - y_{1} = m (x - x_{1})

, the equation of line

N Q

is:

y - 0 = - \frac{4}{3 t} (x - 3 t^{2})

To find the

y

-intercept, we set

x = 0

y = - \frac{4}{3 t} (0 - 3 t^{2}) = 4 t

Given that the

y

-intercept is

\frac{4}{3}

, we solve

4 t = \frac{4}{3}

to find $t = \frac{1}{3}$ .

Final Calculation

Since

M

and

Q

lie on the same horizontal line, the distance

M Q

is the absolute difference of their

x

-coordinates:

M Q = 3 t^{2} - \frac{3 t ^{2}}{4} = \frac{9 t ^{2}}{4}

Substituting

t = \frac{1}{3}

into this expression:

M Q = \frac{9 ( \frac{1}{9} )}{4} = \frac{1}{4}

The final result of our geometric journey is $\frac{1}{4}$ .

Common Traps & FAQs

Why do we use parametric coordinates

(3 t^{2}, 6 t)

instead of

(x, y)

Great question! Using

(x, y)

introduces two variables, which makes the algebra cumbersome. By using the parametric form

P (a t^{2}, 2 a t)

, we reduce the entire problem to a single variable

t

. This is a standard, powerful technique in coordinate geometry that simplifies complex intersections into manageable algebraic expressions.

Why is the line through

M

parallel to the axis simply

y = 3 t

The axis of the parabola

y^{2} = 12 x

is the

x

-axis. A line parallel to the

x

-axis is horizontal, meaning every point on that line has the same

y

-coordinate. Since the line passes through

M (3 t^{2}, 3 t)

, its

y

-coordinate must be

3 t

everywhere. Hence, the equation is

y = 3 t

How do I know if I should use the positive or negative root for

t

In this specific problem, the geometry is symmetric. Whether

t

is positive or negative, the length

M Q

involves

t^{2}

, which will always be positive. The final result for

M Q

will be the same regardless of the sign of

t

. Focus on the magnitude, and you will arrive at the correct answer.

Select Answer:

Visualized Solution

Identify the Parabola

Define Point $P$ Parametrically

Find Foot of Perpendicular $N$

Locate Midpoint $M$

Line Through $M$ Parallel to Axis

Find Coordinates of $Q$

Equation of Line $N Q$

Calculate $y$ -intercept

Solve for Parameter $t$

Calculate Length $M Q$

Conclusion and Verification

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

Analyzing the Setup

Constructing the Geometry

The Intersection of Lines

The Line $N Q$ and the Climax

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Identify the Parabola

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

Find Foot of Perpendicular .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

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

Line Through .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M Parallel to Axis

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

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } NQ

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

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

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

Conclusion and Verification

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

Analyzing the Setup

Constructing the Geometry

The Intersection of Lines

The Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } NQ and the Climax

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line 3x−2y+12=0 intersects the parabola 4y=3x2 at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (x,y) be any point on the parabola y2=4x. Let P be the point that divides the line segment from (0,0) to (x,y) in the ratio 1:3. Then the locus of P is

Define Point $P$ Parametrically

Find Foot of Perpendicular $N$

Locate Midpoint $M$

Line Through $M$ Parallel to Axis

Find Coordinates of $Q$

Equation of Line $N Q$

Calculate $y$ -intercept

Solve for Parameter $t$

Calculate Length $M Q$

The Line $N Q$ and the Climax

If the line $3 x - 2 y + 12 = 0$ intersects the parabola $4 y = 3 x^{2}$ at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is