Is there a constraint on $\theta$ for the normal?

The parameter $\theta$ is the eccentric angle, which can take any value in $[0, 2\pi)$. The optimization holds for all $\theta$ where the normal is defined. The beauty of the trigonometric identity $A^2 \sec^2 \theta + B^2 \csc^2 \theta \ge (A+B)^2$ is that it covers the entire range of $\theta$.

JEE Main 2023 (31 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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:

Q: Wait, why do we minimize the denominator to maximize the distance?

Great question! Look at the distance formula $d = \frac{C}{\sqrt{f(\theta)}}$. Since the numerator $C = a^2 - b^2$ is a constant for a fixed ellipse, the only way to make the fraction $d$ as large as possible is to make the denominator $\sqrt{f(\theta)}$ as small as possible. It is a simple inverse relationship: smaller denominator, larger value!

Q: Why is $a^2 - b^2$ positive in the distance formula?

The distance formula requires the absolute value $|C|$. Since the ellipse is $\frac{x^2}{4} + \frac{y^2}{b^2} = 1$ and we are given $b < 2$, it implies $b^2 < 4$, so $a^2 - b^2 = 4 - b^2$ is indeed positive. We can safely remove the absolute value bars.

Select Answer:

Visualized Solution

Identify the Ellipse Parameters

Given Ellipse: $\frac{x ^{2}}{4} + \frac{y ^{2}}{b ^{2}} = 1$
Comparing with standard form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ :
$a^{2} = 4 ⟹ a = 2$
Constraint: $b < 2$

The Equation of the Normal

General equation of normal at $(a cos θ, b sin θ)$ :
$a x sec θ - b y csc θ = a^{2} - b^{2}$

Substituting Known Values

Substitute $a = 2$ :
$2 x sec θ - b y csc θ = 4 - b^{2}$

Distance from the Origin

Distance $d$ from origin $(0, 0)$ to line $A x + B y + C = 0$ :
$d = \frac{∣ C ∣}{A ^{2} + B ^{2}}$

Applying the Distance Formula

For our normal: $2 x sec θ - b y csc θ - (4 - b^{2}) = 0$
$d = \frac{∣ - ( 4 - b ^{2} ) ∣}{( 2 s e c θ ) ^{2} + ( - b c s c θ ) ^{2}}$
$d = \frac{4 - b ^{2}}{4 s e c ^{2} θ + b ^{2} c s c ^{2} θ}$

Maximizing the Distance

To maximize $d$ , we must minimize the denominator.
Let $f (θ) = 4 sec^{2} θ + b^{2} csc^{2} θ$

Minimum Value of the Denominator

Standard Result: The minimum value of $A^{2} sec^{2} θ + B^{2} csc^{2} θ$ is $(A + B)^{2}$
Here, $A^{2} = 4 ⟹ A = 2$
$B^{2} = b^{2} ⟹ B = b$

Calculating $d_{ma x}$

Minimum value of denominator = $(2 + b)^{2} = 2 + b$
Therefore, $d_{ma x} = \frac{4 - b ^{2}}{2 + b}$
$d_{ma x} = \frac{( 2 - b ) ( 2 + b )}{2 + b} = 2 - b$

Solving for $b$

Given: $d_{ma x} = 1$
$2 - b = 1$
$⟹ b = 1$

Formula for Eccentricity

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

Substituting $a$ and $b$

Substitute $a = 2$ and $b = 1$ :
$e = 1 - \frac{1 ^{2}}{2 ^{2}}$
$e = 1 - \frac{1}{4}$

Final Conclusion

$e = \frac{3}{4} = \frac{3}{2}$
Final Answer: The eccentricity is $\frac{3}{2}$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given ellipse is defined by the equation:

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

Comparing this to the standard form

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

, we identify

a^{2} = 4

, which implies

a = 2

. We are given that

b < 2

, confirming the ellipse is horizontal.

Consider a point

P

on the ellipse with parametric coordinates

(2 cos θ, b sin θ)

. The equation of the normal line at point

P

is given by:

2 x sec θ - b y csc θ = 4 - b^{2}

The Distance Formula

We seek the perpendicular distance

d

from the origin

(0, 0)

to this normal line. Using the standard distance formula

d = \frac{∣ C ∣}{A ^{2} + B ^{2}}

for a line

A x + B y + C = 0

, we substitute our parameters:

d = \frac{∣ - ( 4 - b ^{2} ) ∣}{( 2 sec θ ) ^{2} + ( - b csc θ ) ^{2}}

Since

b < 2

, the term

(4 - b^{2})

is positive. Thus, the distance simplifies to:

d = \frac{4 - b ^{2}}{4 sec ^{2} θ + b ^{2} csc ^{2} θ}

The Optimization Trick

To find the maximum distance

d_{ma x}

, we must minimize the denominator

4 sec^{2} θ + b^{2} csc^{2} θ

. We utilize the identity that the minimum value of

A^{2} sec^{2} θ + B^{2} csc^{2} θ

(A + B)^{2}

Setting

A = 2

and

B = b

, the minimum value of the expression inside the square root is

(2 + b)^{2}

. Taking the square root, the minimum value of the denominator is

2 + b

Substituting this back into our distance expression:

d_{ma x} = \frac{4 - b ^{2}}{2 + b}

Factoring the numerator as

(2 - b) (2 + b)

, the

(2 + b)

terms cancel out, yielding:

d_{ma x} = 2 - b

Final Calculation

The problem states that the maximum distance is

1

. Therefore, we set:

2 - b = 1 \Rightarrow b = 1

With

a = 2

and

b = 1

, we calculate the eccentricity

e

using the formula

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

e = 1 - \frac{1 ^{2}}{2 ^{2}} = 1 - \frac{1}{4} = \frac{3}{4}

The final eccentricity of the ellipse is:

$e = \frac{3}{2}$

Common Traps & FAQs

Wait, why do we minimize the denominator to maximize the distance?

Great question! Look at the distance formula

d = \frac{C}{f ( θ )}

. Since the numerator

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

is a constant for a fixed ellipse, the only way to make the fraction

d

as large as possible is to make the denominator

f (θ)

as small as possible. It is a simple inverse relationship: smaller denominator, larger value!

Is there a constraint on

θ

for the normal?

The parameter

θ

is the eccentric angle, which can take any value in

[0, 2 π)

. The optimization holds for all

θ

where the normal is defined. The beauty of the trigonometric identity

A^{2} sec^{2} θ + B^{2} csc^{2} θ \geq (A + B)^{2}

is that it covers the entire range of

θ

Why is

a^{2} - b^{2}

positive in the distance formula?

The distance formula requires the absolute value

∣ C ∣

. Since the ellipse is

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

and we are given

b < 2

, it implies

b^{2} < 4

, so

a^{2} - b^{2} = 4 - b^{2}

is indeed positive. We can safely remove the absolute value bars.

Animated Solution for Mathematics - Conic Sections: 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:

Select Answer:

Visualized Solution

Identify the Ellipse Parameters

The Equation of the Normal

Substituting Known Values

Distance from the Origin

Applying the Distance Formula

Maximizing the Distance

Minimum Value of the Denominator

Calculating $d_{ma x}$

Solving for $b$

Formula for Eccentricity

Substituting $a$ and $b$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Distance Formula

The Optimization Trick

Final Calculation

Similar Questions

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 :

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

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 :

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 :

Let $P (6, 3)$ be a point on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If the normal at the point $P$ intersects the $x$ -axis at $(9, 0)$ , then the eccentricity of the hyperbola is

Let the eccentricity of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be $\frac{5}{4}$ . If the equation of the normal at the point $(\frac{8}{5}, \frac{12}{5})$ on the hyperbola is $85 x + β y = λ$ , then $λ - β$ is equal to ______.

Select Answer:

Visualized Solution

Identify the Ellipse Parameters

The Equation of the Normal

Substituting Known Values

Distance from the Origin

Applying the Distance Formula

Maximizing the Distance

Minimum Value of the Denominator

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

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

Formula for Eccentricity

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Distance Formula

The Optimization Trick

Final Calculation

Similar Questions

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

.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; } Let P(6,3) be a point on the hyperbola a2x2​−b2y2​=1. If the normal at the point P intersects the x-axis at (9,0), then the eccentricity of the hyperbola is

Calculating $d_{ma x}$

Solving for $b$

Substituting $a$ and $b$

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 :

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 $P (6, 3)$ be a point on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If the normal at the point $P$ intersects the $x$ -axis at $(9, 0)$ , then the eccentricity of the hyperbola is