JEE Main 2020 (4 Sep Evening)

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Animated Solution for Mathematics - Conic Sections: Let $x = 4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$ . If $P (1, β), β > 0$ is a point on this ellipse, then the equation of the normal to it at $P$ is

Q: Why did we reject the negative root for $\beta$?

Great catch on the details! The problem explicitly stated $\beta > 0$. In JEE Advanced, constraints are not just suggestions; they are part of the problem's definition. Always check the domain constraints before proceeding, as they often filter out extraneous solutions.

Select Answer:

Progressive Hints

Hint 1

The directrix is not just a line; it is the fundamental boundary condition that dictates the ellipse's eccentricity. Remember the relationship between the directrix

x = \frac{a}{e}

, the semi-major axis

a

, and the eccentricity

e

. Use this to unlock the value of

a

Hint 2

Once you have the parameters

a

and

b

, you are essentially defining the ellipse's 'DNA'. Use the relation

b^{2} = a^{2} (1 - e^{2})

to find the semi-minor axis. Then, ensure the point

P (1, β)

satisfies the standard equation

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

to find the value of

β

Hint 3

To find the normal, do not waste time deriving the slope from scratch. Use the standard normal equation for an ellipse at point

(x_{1}, y_{1})

\frac{a ^{2} x}{x _{1}} - \frac{b ^{2} y}{y _{1}} = a^{2} - b^{2}

. Simply substitute your known values for

a^{2}, b^{2}, x_{1},

and

y_{1}

to arrive at the final linear equation.

Visualized Solution

Visualizing the Setup

Center: $(0, 0)$
Directrix: $x = 4$
Eccentricity: $e = \frac{1}{2}$

Relating Directrix and Eccentricity

Equation of directrix: $x = \frac{a}{e}$
Given: $\frac{a}{e} = 4$

Finding the Semi-major Axis $a$

Substitute $e = \frac{1}{2}$ :
$\frac{a}{\frac{1}{2}} = 4 \Rightarrow 2 a = 4$
$a = 2$

The Relation for $b^{2}$

Relation: $b^{2} = a^{2} (1 - e^{2})$

Calculating $b^{2}$

$b^{2} = 2^{2} (1 - (\frac{1}{2})^{2})$
$b^{2} = 4 (1 - \frac{1}{4}) = 4 (\frac{3}{4})$
$b^{2} = 3$

Defining the Ellipse Equation

Equation of Ellipse: $\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1$

Point $P (1, β)$ on the Ellipse

Point $P (1, β)$ satisfies $\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1$ :
$\frac{1 ^{2}}{4} + \frac{β ^{2}}{3} = 1$

Solving for $β$

$\frac{β ^{2}}{3} = 1 - \frac{1}{4} = \frac{3}{4}$
$β^{2} = \frac{9}{4} \Rightarrow β = \frac{3}{2}$ (as $β > 0$ )
Point $P = (1, \frac{3}{2})$

The Normal Equation Formula

Normal at $(x_{1}, y_{1})$ is $\frac{a ^{2} x}{x _{1}} - \frac{b ^{2} y}{y _{1}} = a^{2} - b^{2}$

Substituting the Values

Substitute $a^{2} = 4, b^{2} = 3, x_{1} = 1, y_{1} = \frac{3}{2}$ :
$\frac{4 x}{1} - \frac{3 y}{\frac{3}{2}} = 4 - 3$

Final Simplification

$4 x - 3 y \times \frac{2}{3} = 1$
$4 x - 2 y = 1$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Decoding the DNA

We are given the directrix

x = 4

and the eccentricity

e = \frac{1}{2}

. The relationship between the directrix and the semi-major axis

a

is defined by the equation:

x = \frac{a}{e}

By setting

\frac{a}{e} = 4

and substituting our known eccentricity

e = \frac{1}{2}

, we find:

\frac{a}{1/2} = 4 \Rightarrow a = 2

This value of

a = 2

serves as the first pillar of our solution.

The Hidden Geometry

To find the semi-minor axis

b

, we utilize the fundamental relationship between the axes and eccentricity:

b^{2} = a^{2} (1 - e^{2})

Plugging in our known values, we calculate:

b^{2} = 2^{2} (1 - (\frac{1}{2})^{2}) = 4 (1 - \frac{1}{4}) = 4 (\frac{3}{4}) = 3

With

a^{2} = 4

and

b^{2} = 3

, the equation of our ellipse is:

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

The Point of Contact

The point

P (1, β)

lies on the ellipse, meaning it must satisfy the equation derived above. Substituting

x = 1

and

y = β

\frac{1 ^{2}}{4} + \frac{β ^{2}}{3} = 1

Solving for

β^{2}

\frac{β ^{2}}{3} = 1 - \frac{1}{4} = \frac{3}{4} \Rightarrow β^{2} = \frac{9}{4}

Given the constraint

β > 0

, we discard the negative root to obtain

β = \frac{3}{2}

. Thus, the point is

P (1, \frac{3}{2})

The Normal Equation

The standard formula for the equation of the normal at a point

(x_{1}, y_{1})

on an ellipse is:

\frac{a ^{2} x}{x _{1}} - \frac{b ^{2} y}{y _{1}} = a^{2} - b^{2}

Substituting

a^{2} = 4

b^{2} = 3

x_{1} = 1

, and

y_{1} = \frac{3}{2}

, we get:

\frac{4 x}{1} - \frac{3 y}{3/2} = 4 - 3

Simplifying the

y

term, where

\frac{3 y}{3/2} = 3 y \times \frac{2}{3} = 2 y

, we arrive at the final equation:

$4 x - 2 y = 1$

Common Traps & FAQs

Wait, why is the directrix

x = \frac{a}{e}

? I always get confused between the ellipse and the hyperbola.

That is a classic point of friction! Remember, for an ellipse, the directrix is 'outside' the curve, so

x = \frac{a}{e}

is correct because

e < 1

, making

\frac{a}{e} > a

. For a hyperbola,

e > 1

, so

\frac{a}{e} < a

. Keep this physical intuition in mind, and you will never mix them up again.

Why did we reject the negative root for

β

Great catch on the details! The problem explicitly stated

β > 0

. In JEE Advanced, constraints are not just suggestions; they are part of the problem's definition. Always check the domain constraints before proceeding, as they often filter out extraneous solutions.

Is there a faster way to find the normal than the formula?

You could use calculus to find the slope of the tangent and then the negative reciprocal for the normal, but that is a long road. The formula

\frac{a ^{2} x}{x _{1}} - \frac{b ^{2} y}{y _{1}} = a^{2} - b^{2}

is a powerful tool in your arsenal. Memorize it, and you save precious minutes for the tougher problems.

Animated Solution for Mathematics - Conic Sections: Let $x = 4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$ . If $P (1, β), β > 0$ is a point on this ellipse, then the equation of the normal to it at $P$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Relating Directrix and Eccentricity

Finding the Semi-major Axis $a$

The Relation for $b^{2}$

Calculating $b^{2}$

Defining the Ellipse Equation

Point $P (1, β)$ on the Ellipse

Solving for $β$

The Normal Equation Formula

Substituting the Values

Final Simplification

The Sigma Insight: Equation of Tangent and Normal

Decoding the DNA

The Hidden Geometry

The Point of Contact

The Normal Equation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Setup

Relating Directrix and Eccentricity

Finding the Semi-major Axis .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

The Relation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Defining the Ellipse Equation

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(1,β) on the Ellipse

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

The Normal Equation Formula

Substituting the Values

Final Simplification

The Sigma Insight: Equation of Tangent and Normal

Decoding the DNA

The Hidden Geometry

The Point of Contact

The Normal Equation

Similar Questions

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Finding the Semi-major Axis $a$

The Relation for $b^{2}$

Calculating $b^{2}$

Point $P (1, β)$ on the Ellipse

Solving for $β$

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If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point P on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

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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 $α$ .