JEE Main 2020 (6 Sep Evening)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies :

Q: Why did we reject the negative root when we wrote $\frac{b}{a} = \sqrt{1 - e^2}$?

Great catch! In the context of an ellipse, $a$ and $b$ are lengths of the semi-major and semi-minor axes, respectively. By definition, lengths are always positive. Therefore, the ratio $b/a$ must be positive. Since the square root function (the principal root) is defined as positive, $\sqrt{1 - e^2}$ is the correct positive value.

Select Answer:

Visualized Solution

Define the Ellipse and Key Points

Standard Ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
End of Latus Rectum ( $L$ ): $(a e, \frac{b ^{2}}{a})$
Extremity of Minor Axis ( $B^{'}$ ): $(0, - b)$

Equation of the Normal

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

Substitution of Point $L$

Substitute $x_{1} = a e$ and $y_{1} = \frac{b ^{2}}{a}$ :
$\frac{a ^{2} x}{a e} - \frac{b ^{2} y}{b ^{2} / a} = a^{2} - b^{2}$

Simplifying the Normal Equation

Cancel out common terms:
$\frac{a x}{e} - a y = a^{2} - b^{2}$

Applying the Given Condition

The normal passes through $B^{'} (0, - b)$ .
Substitute $x = 0$ and $y = - b$ .

Substituting $B^{'}$ into Normal

$\frac{a ( 0 )}{e} - a (- b) = a^{2} - b^{2}$

Simplifying the Relation

$0 + ab = a^{2} - b^{2}$
Result: $ab = a^{2} - b^{2}$

Relating to Eccentricity

We need an equation in terms of $e$ .
Recall the eccentricity formula: $b^{2} = a^{2} (1 - e^{2})$
Rearranged: $\frac{b ^{2}}{a ^{2}} = 1 - e^{2}$

Dividing by $a^{2}$

Divide $ab = a^{2} - b^{2}$ by $a^{2}$ :
$\frac{ab}{a ^{2}} = \frac{a ^{2}}{a ^{2}} - \frac{b ^{2}}{a ^{2}}$
$\frac{b}{a} = 1 - \frac{b ^{2}}{a ^{2}}$

Substituting Eccentricity

Substitute $\frac{b ^{2}}{a ^{2}} = 1 - e^{2}$
And $\frac{b}{a} = 1 - e^{2}$
Equation becomes: $1 - e^{2} = e^{2}$

Squaring Both Sides

Square both sides to remove the radical:
$(1 - e^{2})^{2} = (e^{2})^{2}$
$1 - e^{2} = e^{4}$

Final Polynomial Equation

Rearrange $1 - e^{2} = e^{4}$ :
$e^{4} + e^{2} - 1 = 0$
This is the required condition for the eccentricity.

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We define our arena using the standard ellipse equation:

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

The latus rectum is a vertical line at

x = a e

. The upper end of this latus rectum, denoted as

L

, has the coordinates:

L = (a e, \frac{b ^{2}}{a})

We also identify the extremity of the minor axis,

B^{'}

, which is located at

(0, - b)

. These points serve as our geometric anchors.

The Normal as a Scalpel

To construct the normal at point

L

, we utilize the general equation of the normal to an ellipse at any point

(x_{1}, y_{1})

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

Substituting our point

L (a e, \frac{b ^{2}}{a})

into this equation, where

x_{1} = a e

and

y_{1} = \frac{b ^{2}}{a}

, we obtain:

\frac{a ^{2} x}{a e} - \frac{b ^{2} y}{b ^{2} / a} = a^{2} - b^{2}

Simplifying the terms, the expression reduces to:

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

The Intersection

The problem states that this normal line passes through the extremity of the minor axis,

B^{'} (0, - b)

. Therefore, the point

(0, - b)

must satisfy the equation of the normal.

Substituting

x = 0

and

y = - b

into the normal equation:

\frac{a ( 0 )}{e} - a (- b) = a^{2} - b^{2}

This simplifies to the elegant relation:

ab = a^{2} - b^{2}

The Bridge to Eccentricity

We recall the fundamental identity for an ellipse,

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

, which implies

\frac{b ^{2}}{a ^{2}} = 1 - e^{2}

. Dividing our relation

ab = a^{2} - b^{2}

a^{2}

, we get:

\frac{b}{a} = 1 - \frac{b ^{2}}{a ^{2}}

Substituting

\frac{b}{a} = 1 - e^{2}

and

\frac{b ^{2}}{a ^{2}} = 1 - e^{2}

into the equation, we obtain:

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

Squaring both sides to eliminate the radical yields

1 - e^{2} = e^{4}

. Rearranging the terms, we arrive at the final condition:

$e^{4} + e^{2} - 1 = 0$

Common Traps & FAQs

Wait, why do we use the normal equation

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

? Can't we just find the slope of the tangent and take the negative reciprocal?

You absolutely can! That is the fundamental way to derive it. However, in JEE Advanced, time is currency. Memorizing the standard normal equation saves you precious minutes. The slope of the tangent at

(x_{1}, y_{1})

- b^{2} x_{1} / a^{2} y_{1}

. The normal slope is the negative reciprocal:

a^{2} y_{1} / b^{2} x_{1}

. Using point-slope form

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

will eventually lead you to the same result, but the standard form is cleaner and less prone to algebraic slips.

Why did we reject the negative root when we wrote

\frac{b}{a} = 1 - e^{2}

Great catch! In the context of an ellipse,

a

and

b

are lengths of the semi-major and semi-minor axes, respectively. By definition, lengths are always positive. Therefore, the ratio

b / a

must be positive. Since the square root function (the principal root) is defined as positive,

1 - e^{2}

is the correct positive value.

Is there a geometric shortcut to this problem without doing all this algebra?

While there are properties of the ellipse that relate the latus rectum to the axes, deriving the condition algebraically is the most robust method. JEE Advanced problems often test your ability to manipulate algebraic expressions involving

e

. Trust the process—the algebra here is elegant and leads directly to the answer.

Animated Solution for Mathematics - Conic Sections: If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies :

Select Answer:

Visualized Solution

Define the Ellipse and Key Points

Equation of the Normal

Substitution of Point $L$

Simplifying the Normal Equation

Applying the Given Condition

Substituting $B^{'}$ into Normal

Simplifying the Relation

Relating to Eccentricity

Dividing by $a^{2}$

Substituting Eccentricity

Squaring Both Sides

Final Polynomial Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal as a Scalpel

The Intersection

The Bridge to Eccentricity

Similar Questions

If the maximum distance of normal to the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{b ^{2}} = 1, b < 2$ , from the origin is 1, then the eccentricity of the ellipse is:

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directrices is $x = - 4$ , then the equation of the normal to it at $(1, \frac{3}{2})$ is:

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

The length of minor axis (along y-axis) of an ellipse of the standard form is $4/ 3$ . If this ellipse touches the line $x + 6 y = 8$ , then its eccentricity is :

The length of the minor axis (along y-axis) of an ellipse in the standard form is $\frac{4}{3}$ . If this ellipse touches the line, $x + 6 y = 8$ ; then its eccentricity is :

Let the line $y = m x$ and the ellipse $2 x^{2} + y^{2} = 1$ intersect a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(- \frac{1}{3 2}, 0)$ and $(0, β)$ , then $β$ is equal to :

The normal at a point $P$ on the ellipse $x^{2} + 4 y^{2} = 16$ meets the $x$ -axis at $Q$ . If $M$ is the mid point of the line segment $P Q$ , then the locus of $M$ intersects the latus rectums of the given ellipse at the points

Let the line $y = m x$ and the ellipse $2 x^{2} + y^{2} = 1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(\frac{1}{3 2}, 0)$ and $(0, β)$ , then $β$ is equal to :

Select Answer:

Visualized Solution

Define the Ellipse and Key Points

Equation of the Normal

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

Simplifying the Normal Equation

Applying the Given Condition

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B′ into Normal

Simplifying the Relation

Relating to Eccentricity

Dividing by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Substituting Eccentricity

Squaring Both Sides

Final Polynomial Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal as a Scalpel

The Intersection

The Bridge to Eccentricity

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the maximum distance of normal to the ellipse 4x2​+b2y2​=1,b<2, from the origin is 1, then the eccentricity of the ellipse is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The eccentricity of an ellipse whose centre is at the origin is 21​. If one of its directrices is x=−4, then the equation of the normal to it at (1,23​) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of minor axis (along y-axis) of an ellipse of the standard form is 4/3​. If this ellipse touches the line x+6y=8, then its eccentricity is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the minor axis (along y-axis) of an ellipse in the standard form is 3​4​. If this ellipse touches the line, x+6y=8; then its eccentricity is :

Substitution of Point $L$

Substituting $B^{'}$ into Normal

Dividing by $a^{2}$

If the maximum distance of normal to the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{b ^{2}} = 1, b < 2$ , from the origin is 1, then the eccentricity of the ellipse is:

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directrices is $x = - 4$ , then the equation of the normal to it at $(1, \frac{3}{2})$ is:

The length of minor axis (along y-axis) of an ellipse of the standard form is $4/ 3$ . If this ellipse touches the line $x + 6 y = 8$ , then its eccentricity is :

The length of the minor axis (along y-axis) of an ellipse in the standard form is $\frac{4}{3}$ . If this ellipse touches the line, $x + 6 y = 8$ ; then its eccentricity is :