JEE Main 2020 (8 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let a line $y = m x (m > 0)$ intersect the parabola, $y^{2} = x$ at a point $P$ , other than the origin. Let the tangent to it at $P$ meet the $x$ -axis at the point $Q$ . If area $(Δ O P Q) = 4$ sq. units, then $m$ is equal to _________ .

Q: Wait, why do we use $t = 1/m$ instead of just solving for $m$ directly?

Great question! While you absolutely could solve for $m$ directly, it leads to expressions involving $1/m^2$ and $1/m$, which are prone to algebraic errors. By substituting $t = 1/m$, we transform the coordinates into the elegant parametric form $(t^2, t)$. It is a classic trick in coordinate geometry to simplify the math and keep your focus on the concepts rather than the arithmetic.

Q: Can we use the formula for the tangent in terms of slope $m$?

Yes, you can! The tangent to $y^2 = 4ax$ with slope $m$ is $y = mx + a/m$. Here, $4a = 1$, so $a = 1/4$. The tangent is $y = mx + 1/(4m)$. This is a perfectly valid approach. However, the parametric method is often more intuitive when dealing with specific points like $P$.

Q: Why is the area of the triangle $\frac{1}{2} t^3$?

Think of the triangle $\Delta OPQ$ with vertices $O(0,0)$, $P(t^2, t)$, and $Q(-t^2, 0)$. The base lies on the $x$-axis from $x = -t^2$ to $x = 0$, so the length of the base is $t^2$. The height is the vertical distance of point $P$ from the $x$-axis, which is the $y$-coordinate of $P$, namely $t$. The area is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t^2 \times t = \frac{1}{2} t^3$. It is a beautiful, simple result!

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

Given Parabola: $y^{2} = x$
Given Line: $y = m x$ where $m > 0$
Point $P$ is the intersection of the line and the parabola (other than origin).

Finding Point $P$

Substitute $y = m x$ into $y^{2} = x$
$(m x)^{2} = x \Rightarrow m^{2} x^{2} = x$
$x (m^{2} x - 1) = 0$

Coordinates of $P$

For $P \neq = (0, 0)$ , $x = \frac{1}{m ^{2}}$
Substitute $x$ back into $y = m x$ :
$y = m (\frac{1}{m ^{2}}) = \frac{1}{m}$
Point $P = (\frac{1}{m ^{2}}, \frac{1}{m})$

Introducing Parameter $t$

Let $t = \frac{1}{m}$
Then $P = (t^{2}, t)$ where $t > 0$
This simplifies future calculations.

Equation of Tangent at $P$

Equation of tangent at $(x_{1}, y_{1})$ to $y^{2} = x$ is $y y_{1} = \frac{1}{2} (x + x_{1})$
At $P (t^{2}, t)$ , substitute $x_{1} = t^{2}$ and $y_{1} = t$
Tangent equation: $y t = \frac{1}{2} (x + t^{2})$

Finding Point $Q$

The tangent meets the x-axis at point $Q$ .
On the x-axis, $y = 0$ .
Substitute $y = 0$ into $y t = \frac{1}{2} (x + t^{2})$
$0 = \frac{1}{2} (x + t^{2}) \Rightarrow x = - t^{2}$
Point $Q = (- t^{2}, 0)$

Visualizing Triangle $O P Q$

Vertices of $Δ O P Q$ :
$O (0, 0)$
$P (t^{2}, t)$
$Q (- t^{2}, 0)$

Area Formula Setup

Area of $Δ = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$
Substitute $O (0, 0), P (t^{2}, t), Q (- t^{2}, 0)$ :
Area $= \frac{1}{2} 0 (t - 0) + t^{2} (0 - 0) + (- t^{2}) (0 - t)$

Calculating the Area

Simplify the expression:
Area $= \frac{1}{2} 0 + 0 + (- t^{2}) (- t)$
Area $= \frac{1}{2} t^{3}$
Since $t > 0$ , Area $= \frac{1}{2} t^{3}$

Equating to Given Area

We are given that Area $= 4$ sq. units.
Equate our expression to 4:
$\frac{1}{2} t^{3} = 4$
Multiply both sides by 2:
$t^{3} = 8$

Solving for $t$

$t^{3} = 8$
We know that $2^{3} = 8$
Taking the real cube root on both sides:
$t = 2$

Finding the Slope $m$

Recall our initial substitution: $t = \frac{1}{m}$
Rearranging for $m$ : $m = \frac{1}{t}$
Substitute $t = 2$ :
$m = \frac{1}{2} = 0.5$
Final Answer: $m = 0.5$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Intersection

To find point

P

, we solve the system

y^{2} = x

and

y = m x

. Substituting

y = m x

into the parabola equation gives

(m x)^{2} = x

, which simplifies to

m^{2} x^{2} - x = 0

Factoring this, we get

x (m^{2} x - 1) = 0

. Since we are looking for a point other than the origin, we ignore

x = 0

and find

x = 1/ m^{2}

Consequently,

y = m (1/ m^{2}) = 1/ m

. So, the point of intersection is

P = (1/ m^{2}, 1/ m)

To simplify our calculations, let us define a parameter

t = 1/ m

. Now, our point

P

becomes

(t^{2}, t)

The Tangent's Geometry

Now, we draw a tangent at

P (t^{2}, t)

. The standard equation for a tangent to

y^{2} = x

(x_{1}, y_{1})

is given by:

y y_{1} = \frac{1}{2} (x + x_{1})

Substituting our coordinates, we get

y t = \frac{1}{2} (x + t^{2})

. This line intersects the

x

-axis at point

Q

At the

x

-axis,

y = 0

, so

0 = \frac{1}{2} (x + t^{2})

, which means

x = - t^{2}

. Thus,

Q = (- t^{2}, 0)

The Area Calculation

The area of a triangle with vertices

(x_{1}, y_{1})

(x_{2}, y_{2})

, and

(x_{3}, y_{3})

is given by:

Area = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣

Plugging in our points

O (0, 0)

P (t^{2}, t)

, and

Q (- t^{2}, 0)

, the formula simplifies significantly. The first two terms vanish because of the origin, leaving us with:

Area = \frac{1}{2} ∣ (- t^{2}) (0 - t) ∣ = \frac{1}{2} ∣ t^{3} ∣

Since

t > 0

, the area is simply

\frac{1}{2} t^{3}

The Final Reveal

We are given that the area of the triangle is

4

. Therefore, we set up the equation:

\frac{1}{2} t^{3} = 4 \Rightarrow t^{3} = 8

Taking the cube root, we find

t = 2

. Since we defined

t = 1/ m

, we have

2 = 1/ m

This leads us to the final result:

$m = \frac{1}{2}$

Common Traps & FAQs

Wait, why do we use

t = 1/ m

instead of just solving for

m

directly?

Great question! While you absolutely could solve for

m

directly, it leads to expressions involving

1/ m^{2}

and

1/ m

, which are prone to algebraic errors. By substituting

t = 1/ m

, we transform the coordinates into the elegant parametric form

(t^{2}, t)

. It is a classic trick in coordinate geometry to simplify the math and keep your focus on the concepts rather than the arithmetic.

Can we use the formula for the tangent in terms of slope

m

Yes, you can! The tangent to

y^{2} = 4 a x

with slope

m

y = m x + a / m

. Here,

4 a = 1

, so

a = 1/4

. The tangent is

y = m x + 1/ (4 m)

. This is a perfectly valid approach. However, the parametric method is often more intuitive when dealing with specific points like

P

Why is the area of the triangle

\frac{1}{2} t^{3}

Think of the triangle

Δ O P Q

with vertices

O (0, 0)

P (t^{2}, t)

, and

Q (- t^{2}, 0)

. The base lies on the

x

-axis from

x = - t^{2}

x = 0

, so the length of the base is

t^{2}

. The height is the vertical distance of point

P

from the

x

-axis, which is the

y

-coordinate of

P

, namely

t

. The area is

\frac{1}{2} \times base \times height = \frac{1}{2} \times t^{2} \times t = \frac{1}{2} t^{3}

. It is a beautiful, simple result!

Animated Solution for Mathematics - Conic Sections: Let a line $y = m x (m > 0)$ intersect the parabola, $y^{2} = x$ at a point $P$ , other than the origin. Let the tangent to it at $P$ meet the $x$ -axis at the point $Q$ . If area $(Δ O P Q) = 4$ sq. units, then $m$ is equal to _________ .

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

Finding Point $P$

Coordinates of $P$

Introducing Parameter $t$

Equation of Tangent at $P$

Finding Point $Q$

Visualizing Triangle $O P Q$

Area Formula Setup

Calculating the Area

Equating to Given Area

Solving for $t$

Finding the Slope $m$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Intersection

The Tangent's Geometry

The Area Calculation

The Final Reveal

Similar Questions

Let a line $y = m x (m > 0)$ intersect the parabola, $y^{2} = x$ at a point $P$ , other than the origin. Let the tangent to it at $P$ meet the $x$ -axis at the point $Q$ . If area $(Δ O P Q) = 4$ sq. units, then $m$ is equal to

If $y = m x + 4$ is a tangent to both the parabolas $y^{2} = 4 x$ and $x^{2} = 2 b y$ . Then value of $b$ is

Let a parabola $P$ be such that its vertex and focus lie on the positive $x$ -axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from $O (0, 0)$ to the parabola $P$ which meet $P$ at $S$ and $R$ , then the area (in sq. units) of $Δ S O R$ is equal to

If $y = m x + 4$ is a tangent to both the parabolas, $y^{2} = 4 x$ and $x^{2} = 2 b y$ , then $b$ is equal to

Normals are drawn from the point $P$ with slopes $m_{1}, m_{2}, m_{3}$ to the parabola $y^{2} = 4 x$ . If locus of $P$ with $m_{1} m_{2} = α$ is a part of the parabola itself then find $α$ .

If $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}, m_{1} \neq = m_{2}$ are two common tangents of circle $x^{2} + y^{2} = 2$ and parabola $y^{2} = x$ , then the value of $8∣ m_{1} m_{2} ∣$ is equal to

If the line $y = 4 + k x, k > 0$ , is the tangent to the parabola $y = x - x^{2}$ at the point $P$ and $V$ is the vertex of the parabola, then the slope of the line through $P$ and $V$ is :

Let the tangent to the parabola $S : y^{2} = 2 x$ at the point $P (2, 2)$ meet the $x$ -axis at $Q$ and normal at it meet the parabola $S$ at the point $R$ . Then the area (in sq. units) of the triangle $P QR$ is equal to:

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

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

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

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

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

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

Visualizing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OPQ

Area Formula Setup

Calculating the Area

Equating to Given Area

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

Finding the Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Intersection

The Tangent's Geometry

The Area Calculation

The Final Reveal

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=mx+4 is a tangent to both the parabolas, y2=4x and x2=2by, then b is equal to

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Finding Point $P$

Coordinates of $P$

Introducing Parameter $t$

Equation of Tangent at $P$

Finding Point $Q$

Visualizing Triangle $O P Q$

Solving for $t$

Finding the Slope $m$

If $y = m x + 4$ is a tangent to both the parabolas $y^{2} = 4 x$ and $x^{2} = 2 b y$ . Then value of $b$ is

If $y = m x + 4$ is a tangent to both the parabolas, $y^{2} = 4 x$ and $x^{2} = 2 b y$ , then $b$ is equal to

Normals are drawn from the point $P$ with slopes $m_{1}, m_{2}, m_{3}$ to the parabola $y^{2} = 4 x$ . If locus of $P$ with $m_{1} m_{2} = α$ is a part of the parabola itself then find $α$ .

If $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}, m_{1} \neq = m_{2}$ are two common tangents of circle $x^{2} + y^{2} = 2$ and parabola $y^{2} = x$ , then the value of $8∣ m_{1} m_{2} ∣$ is equal to

If the line $y = 4 + k x, k > 0$ , is the tangent to the parabola $y = x - x^{2}$ at the point $P$ and $V$ is the vertex of the parabola, then the slope of the line through $P$ and $V$ is :