JEE Main 2020 (9 Jan Evening)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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 :

Q: Wait, why is the minor axis $2b$ and not $2a$?

Great question! In the standard ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the length of the axis along the x-axis is $2a$ and along the y-axis is $2b$. Since the problem specifies the minor axis is along the y-axis, we know $b < a$, so the length of the minor axis is indeed $2b$. Always check the orientation!

Q: Can I use the discriminant method instead of the tangency condition?

Absolutely! You can substitute $y = \frac{8-x}{6}$ into the ellipse equation, form a quadratic in $x$, and set the discriminant $D = 0$ for tangency. However, the condition $c^2 = a^2m^2 + b^2$ is a derived shortcut that saves significant time in JEE exams. Both methods are mathematically sound.

Select Answer:

Visualized Solution

Standard Ellipse Setup

Let the standard ellipse be $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
The minor axis lies along the y-axis.

Minor Axis Length

Length of minor axis $= 2 b = \frac{4}{3}$
Therefore, $b = \frac{2}{3}$

Calculate $b^{2}$

Squaring both sides:
$b^{2} = (\frac{2}{3})^{2} = \frac{4}{3}$

The Tangent Line

The ellipse touches the line: $x + 6 y = 8$
This line is a tangent to the ellipse.

Slope-Intercept Form

Convert $x + 6 y = 8$ to $y = m x + c$
$6 y = - x + 8$
$y = - \frac{1}{6} x + \frac{8}{6}$

Extract $m$ and $c$

$y = - \frac{1}{6} x + \frac{4}{3}$
Slope $m = - \frac{1}{6}$
y-intercept $c = \frac{4}{3}$

Condition for Tangency

For a line $y = m x + c$ to touch $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ :
$c^{2} = a^{2} m^{2} + b^{2}$

Substitute Known Values

Substitute $c = \frac{4}{3}$ , $m = - \frac{1}{6}$ , and $b^{2} = \frac{4}{3}$ :
$(\frac{4}{3})^{2} = a^{2} (- \frac{1}{6})^{2} + \frac{4}{3}$

Expand the Equation

$\frac{16}{9} = a^{2} (\frac{1}{36}) + \frac{4}{3}$
$\frac{16}{9} = \frac{a ^{2}}{36} + \frac{4}{3}$

Isolate $a^{2}$

$\frac{a ^{2}}{36} = \frac{16}{9} - \frac{4}{3}$
$\frac{a ^{2}}{36} = \frac{16 - 12}{9}$
$\frac{a ^{2}}{36} = \frac{4}{9}$

Solve for $a^{2}$

$a^{2} = 36 \times \frac{4}{9}$
$a^{2} = 4 \times 4 = 16$

Eccentricity Formula

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

Substitute into Eccentricity

Substitute $a^{2} = 16$ and $b^{2} = \frac{4}{3}$ :
$e = 1 - \frac{\frac{4}{3}}{16}$

Simplify the Fraction

$e = 1 - \frac{4}{3 \times 16}$
$e = 1 - \frac{1}{12}$

Final Calculation

$e = \frac{12 - 1}{12} = \frac{11}{12}$
$e = \frac{11}{4 \times 3} = \frac{11}{2 3}$
$e = \frac{1}{2} \frac{11}{3}$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The standard form of an ellipse is given by:

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

The problem states that the minor axis lies along the

y

-axis, implying the ellipse is stretched horizontally such that

a > b

. We are given the length of the minor axis as

\frac{4}{3}

Since the length of the minor axis is

2 b

, we have:

2 b = \frac{4}{3} \Rightarrow b = \frac{2}{3}

Squaring this value, we obtain our first vital constant:

b^{2} = \frac{4}{3}

The Tangency Bridge

We are given the line

x + 6 y = 8

, which is tangent to the ellipse. To utilize this, we transform the line into slope-intercept form

y = m x + c

6 y = - x + 8 \Rightarrow y = - \frac{1}{6} x + \frac{4}{3}

Here, the slope is

m = - \frac{1}{6}

and the

y

-intercept is

c = \frac{4}{3}

. The condition for a line

y = m x + c

to be tangent to the ellipse

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

is:

c^{2} = a^{2} m^{2} + b^{2}

The Algebraic Dance

Substituting our known values into the tangency condition:

(\frac{4}{3})^{2} = a^{2} (- \frac{1}{6})^{2} + \frac{4}{3}

Squaring the terms yields:

\frac{16}{9} = a^{2} (\frac{1}{36}) + \frac{4}{3}

To isolate

a^{2}

, we subtract

\frac{4}{3}

(or

\frac{12}{9}

) from both sides:

\frac{16}{9} - \frac{12}{9} = \frac{a ^{2}}{36} \Rightarrow \frac{4}{9} = \frac{a ^{2}}{36}

Multiplying both sides by

36

, we find:

a^{2} = 36 \times \frac{4}{9} = 16

The Final Reveal

With

a^{2} = 16

and

b^{2} = \frac{4}{3}

, we calculate the eccentricity

e

using the formula

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

e = 1 - \frac{4/3}{16} = 1 - \frac{4}{48} = 1 - \frac{1}{12} = \frac{11}{12}

Simplifying the radical expression:

e = \frac{11}{4 \times 3} = \frac{11}{2 3} = \frac{1}{2} \frac{11}{3}

The final eccentricity of the ellipse is $\frac{1}{2} \frac{11}{3}$ .

Common Traps & FAQs

Wait, why is the minor axis

2 b

and not

2 a

Great question! In the standard ellipse equation

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

, the length of the axis along the x-axis is

2 a

and along the y-axis is

2 b

. Since the problem specifies the minor axis is along the y-axis, we know

b < a

, so the length of the minor axis is indeed

2 b

. Always check the orientation!

Can I use the discriminant method instead of the tangency condition?

Absolutely! You can substitute

y = \frac{8 - x}{6}

into the ellipse equation, form a quadratic in

x

, and set the discriminant

D = 0

for tangency. However, the condition

c^{2} = a^{2} m^{2} + b^{2}

is a derived shortcut that saves significant time in JEE exams. Both methods are mathematically sound.

How do I know if the ellipse is horizontal or vertical?

The problem explicitly states the minor axis is along the y-axis. This means the ellipse is 'squashed' vertically and 'stretched' horizontally, making it a horizontal ellipse where

a > b

. If the minor axis were along the x-axis, it would be a vertical ellipse where

b > a

Animated Solution for Mathematics - Conic Sections: 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 :

Select Answer:

Visualized Solution

Standard Ellipse Setup

Minor Axis Length

Calculate $b^{2}$

The Tangent Line

Slope-Intercept Form

Extract $m$ and $c$

Condition for Tangency

Substitute Known Values

Expand the Equation

Isolate $a^{2}$

Solve for $a^{2}$

Eccentricity Formula

Substitute into Eccentricity

Simplify the Fraction

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangency Bridge

The Algebraic Dance

The Final Reveal

Similar Questions

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 :

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:

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 a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285$ . Then the common tangents are inclined to the minor axis of the ellipse at the angle

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:

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ , for some $a \in R$ then the distance between the foci of the ellipse is :

Select Answer:

Visualized Solution

Standard Ellipse Setup

Minor Axis Length

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

The Tangent Line

Slope-Intercept Form

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

Condition for Tangency

Substitute Known Values

Expand the Equation

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

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

Eccentricity Formula

Substitute into Eccentricity

Simplify the Fraction

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangency Bridge

The Algebraic Dance

The Final Reveal

Similar Questions

.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; } 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; } 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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a circle of radius 4 be concentric to the ellipse 15x2+19y2=285. Then the common tangents are inclined to the minor axis of the ellipse at the angle

.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; } If 3x+4y=122​ is a tangent to the ellipse a2x2​+9y2​=1, for some a∈R then the distance between the foci of the ellipse is :

Calculate $b^{2}$

Extract $m$ and $c$

Isolate $a^{2}$

Solve for $a^{2}$

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 :

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:

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 a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285$ . Then the common tangents are inclined to the minor axis of the ellipse at the angle

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:

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ , for some $a \in R$ then the distance between the foci of the ellipse is :