JEE Main 2022 (28 June Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let the eccentricity of the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be $\frac{5}{2}$ and length of its latus rectum be $62$ . If $y = 2 x + c$ is a tangent to the hyperbola $H$ , then the value of $c^{2}$ is equal to:

Q: Wait, why do we use $e^2 = 1 + \frac{b^2}{a^2}$? Isn't it $1 - \frac{b^2}{a^2}$?

That is a classic confusion! The formula $e^2 = 1 - \frac{b^2}{a^2}$ is for an ellipse, where the eccentricity is less than 1. For a hyperbola, the curve is 'flatter' and more open, so the eccentricity must be greater than 1. Thus, the formula is $e^2 = 1 + \frac{b^2}{a^2}$. Always keep the conic section type in mind!

Q: Why is the tangency condition $c^2 = a^2m^2 - b^2$ and not $c^2 = a^2m^2 + b^2$?

Great observation! The condition $c^2 = a^2m^2 + b^2$ applies to an ellipse. For a hyperbola, the negative sign in the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ carries over into the tangency condition, resulting in $c^2 = a^2m^2 - b^2$. It is a direct consequence of the hyperbola's geometry.

Q: Can I just use the discriminant method ($D=0$) to find $c$?

Absolutely! You can substitute $y = 2x + c$ into the hyperbola equation, form a quadratic in $x$, and set the discriminant to zero. However, using the standard tangency condition $c^2 = a^2m^2 - b^2$ is much faster and less prone to algebraic errors in a high-pressure exam like JEE Advanced. Both methods are mathematically sound!

Select Answer:

Visualized Solution

The Hyperbola $H$

Equation: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Eccentricity: $e = \frac{5}{2}$
Latus Rectum: $L R = 62$

Eccentricity Formula

Formula: $e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$

Substituting $e$

Substitute $e = \frac{5}{2}$ :
$(\frac{5}{2})^{2} = 1 + \frac{b ^{2}}{a ^{2}}$
$\frac{5}{2} = 1 + \frac{b ^{2}}{a ^{2}}$

Ratio of $b^{2}$ and $a^{2}$

$\frac{b ^{2}}{a ^{2}} = \frac{5}{2} - 1$
$\frac{b ^{2}}{a ^{2}} = \frac{3}{2}$
$b^{2} = \frac{3 a ^{2}}{2}$

Latus Rectum Formula

Length of Latus Rectum: $L R = \frac{2 b ^{2}}{a}$
Given: $L R = 62$

Substituting $b^{2}$

Substitute $b^{2} = \frac{3 a ^{2}}{2}$ :
$\frac{2 ( \frac{3 a ^{2}}{2} )}{a} = 62$

Solving for $a$

$\frac{3 a ^{2}}{a} = 62$
$3 a = 62$
$a = 22$

Values of $a^{2}$ and $b^{2}$

$a^{2} = (22)^{2} = 8$
$b^{2} = \frac{3}{2} (8) = 12$

Tangency Condition

Tangent line: $y = 2 x + c$
Slope $m = 2$
Condition for tangency: $c^{2} = a^{2} m^{2} - b^{2}$

Calculating $c^{2}$

Substitute $a^{2} = 8$ , $b^{2} = 12$ , $m = 2$ :
$c^{2} = 8 (2)^{2} - 12$
$c^{2} = 8 (4) - 12$

Final Result

$c^{2} = 32 - 12$
$c^{2} = 20$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The hyperbola is defined by the standard equation:

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

We are given the eccentricity

e = \frac{5}{2}

and the length of the latus rectum

L R = 62

. The eccentricity

e

is related to the semi-axes

a

and

b

by the fundamental identity:

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

Substituting the given value of

e

into this identity, we obtain:

(\frac{5}{2})^{2} = 1 + \frac{b ^{2}}{a ^{2}} ⟹ \frac{5}{2} = 1 + \frac{b ^{2}}{a ^{2}}

Subtracting 1 from both sides yields the ratio:

\frac{b ^{2}}{a ^{2}} = \frac{3}{2} ⟹ b^{2} = \frac{3 a ^{2}}{2}

The Latus Rectum

The Width of the Focus

The length of the latus rectum for a hyperbola is given by the formula:

L R = \frac{2 b ^{2}}{a}

Given

L R = 62

, we substitute our expression for

b^{2}

into this formula:

\frac{2 ( \frac{3 a ^{2}}{2} )}{a} = 62

Simplifying the expression by canceling the 2s and dividing by

a

3 a = 62 ⟹ a = 22

Now, we calculate the squares of the semi-axes:

a^{2} = (22)^{2} = 8

b^{2} = \frac{3}{2} (8) = 12

The Tangent's Dance

We consider the line

y = 2 x + c

, which is tangent to the hyperbola. For a line

y = m x + c

to be tangent to the hyperbola

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

, it must satisfy the condition:

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

Here, the slope

m = 2

. Substituting the known values of

a^{2}

b^{2}

, and

m

into the condition:

c^{2} = 8 (2)^{2} - 12

c^{2} = 8 (4) - 12 = 32 - 12

The final value is $c^{2} = 20$ .

Common Traps & FAQs

Wait, why do we use

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

? Isn't it

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

That is a classic confusion! The formula

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

is for an ellipse, where the eccentricity is less than 1. For a hyperbola, the curve is 'flatter' and more open, so the eccentricity must be greater than 1. Thus, the formula is

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

. Always keep the conic section type in mind!

Why is the tangency condition

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

and not

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

Great observation! The condition

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

applies to an ellipse. For a hyperbola, the negative sign in the standard equation

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

carries over into the tangency condition, resulting in

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

. It is a direct consequence of the hyperbola's geometry.

Can I just use the discriminant method (

D = 0

) to find

c

Absolutely! You can substitute

y = 2 x + c

into the hyperbola equation, form a quadratic in

x

, and set the discriminant to zero. However, using the standard tangency condition

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

is much faster and less prone to algebraic errors in a high-pressure exam like JEE Advanced. Both methods are mathematically sound!

Animated Solution for Mathematics - Conic Sections: Let the eccentricity of the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be $\frac{5}{2}$ and length of its latus rectum be $62$ . If $y = 2 x + c$ is a tangent to the hyperbola $H$ , then the value of $c^{2}$ is equal to:

Select Answer:

Visualized Solution

The Hyperbola $H$

Eccentricity Formula

Substituting $e$

Ratio of $b^{2}$ and $a^{2}$

Latus Rectum Formula

Substituting $b^{2}$

Solving for $a$

Values of $a^{2}$ and $b^{2}$

Tangency Condition

Calculating $c^{2}$

Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Latus Rectum

The Tangent's Dance

Similar Questions

If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is-

The point $P (- 26, 3)$ lies on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ having eccentricity $\frac{5}{2}$ . If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then QR is equal to :

If the line $y = m x + c$ is a common tangent to the hyperbola $\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1$ and the circle $x^{2} + y^{2} = 36$ , then which one of the following is true?

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 ______.

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

A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point:

Select Answer:

Visualized Solution

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

Eccentricity Formula

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

Ratio of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Latus Rectum Formula

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

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

Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Tangency Condition

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c2

Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Latus Rectum

The Tangent's Dance

Similar Questions

If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line y=mx+c is a common tangent to the hyperbola 100x2​−64y2​=1 and the circle x2+y2=36, then which one of the following is true?

.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; } A hyperbola passes through the point P(2​,3​) and has foci at (±2,0). Then the tangent to this hyperbola at P also passes through the point:

The Hyperbola $H$

Substituting $e$

Ratio of $b^{2}$ and $a^{2}$

Substituting $b^{2}$

Solving for $a$

Values of $a^{2}$ and $b^{2}$

Calculating $c^{2}$

If the line $y = m x + c$ is a common tangent to the hyperbola $\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1$ and the circle $x^{2} + y^{2} = 36$ , then which one of the following is true?

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

A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point: