JEE Advanced 2011

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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

Q: Why is the normal equation $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$ and not just the slope-point form?

You absolutely can use the slope-point form $y - y_1 = m(x - x_1)$, but it requires you to first find the slope of the tangent, then take its negative reciprocal to find the slope of the normal. The equation $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$ is simply the derived, optimized form for the normal to a hyperbola. Using it saves time and reduces the chance of algebraic errors in the heat of an exam.

Q: Wait, why does the $y$-term vanish when we substitute $(9, 0)$?

Great observation! The point $(9, 0)$ lies on the $x$-axis, meaning its $y$-coordinate is $0$. When you substitute $y = 0$ into the equation of the normal, the term $\frac{b^2y}{y_1}$ becomes $\frac{b^2(0)}{3}$, which is $0$. This is a gift from the problem setter—it simplifies the equation significantly, allowing us to isolate the relationship between $a^2$ and $b^2$.

Select Answer:

Visualized Solution

Visualizing the Hyperbola and Point $P$

Consider the standard hyperbola equation: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
A point $P (6, 3)$ lies on this hyperbola.
We draw a normal at $P$ which intersects the $x$ -axis at $(9, 0)$ .

Point $P (6, 3)$ on the Hyperbola

Since $P (6, 3)$ lies on the hyperbola, it must satisfy its equation.
Substituting $x = 6$ and $y = 3$ gives:
$\frac{6 ^{2}}{a ^{2}} - \frac{3 ^{2}}{b ^{2}} = 1 ⟹ \frac{36}{a ^{2}} - \frac{9}{b ^{2}} = 1$

Equation of the Normal

The general equation of the normal to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ at any point $(x_{1}, y_{1})$ is:
$\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}$

Substituting Point $P (6, 3)$

Substitute $(x_{1}, y_{1}) = (6, 3)$ into the normal equation:
$\frac{a ^{2} x}{6} + \frac{b ^{2} y}{3} = a^{2} + b^{2}$

Using the $x$ -axis Intersection

The normal intersects the $x$ -axis at $(9, 0)$ .
Substitute $x = 9$ and $y = 0$ into the equation:
$\frac{a ^{2} ( 9 )}{6} + \frac{b ^{2} ( 0 )}{3} = a^{2} + b^{2}$

Simplifying the Equation

The term containing $b^{2}$ becomes zero.
The equation simplifies to:
$\frac{3}{2} a^{2} = a^{2} + b^{2}$

Finding the Relation Between $a^{2}$ and $b^{2}$

Subtract $a^{2}$ from both sides:
$\frac{3}{2} a^{2} - a^{2} = b^{2} ⟹ \frac{1}{2} a^{2} = b^{2}$
This gives the ratio: $\frac{b ^{2}}{a ^{2}} = \frac{1}{2}$

Formula for Eccentricity $e$

The eccentricity $e$ of a hyperbola is given by:
$e = 1 + \frac{b ^{2}}{a ^{2}}$

Calculating Eccentricity $e$

Substitute $\frac{b ^{2}}{a ^{2}} = \frac{1}{2}$ into the formula:
$e = 1 + \frac{1}{2} = \frac{3}{2}$

Final Conclusion

The eccentricity of the hyperbola is $\frac{3}{2}$ .
This matches Option 2.

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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 that the point

P (6, 3)

lies on this curve. This implies that the coordinates must satisfy the hyperbola's equation.

Substituting

x = 6

and

y = 3

into the equation, we obtain:

\frac{36}{a ^{2}} - \frac{9}{b ^{2}} = 1

This serves as our first fundamental relationship between the semi-axes

a

and

b

The Normal Line Equation

The equation of the normal to the hyperbola

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

at a specific point

(x_{1}, y_{1})

is given by:

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

Substituting the coordinates of point

P (6, 3)

into this formula, we define the specific normal line for our hyperbola:

\frac{a ^{2} x}{6} + \frac{b ^{2} y}{3} = a^{2} + b^{2}

The Intersection Constraint

We are given that this normal line intersects the

x

-axis at the point

(9, 0)

. By substituting

x = 9

and

y = 0

into the normal equation, the

y

-term vanishes:

\frac{a ^{2} ( 9 )}{6} + \frac{b ^{2} ( 0 )}{3} = a^{2} + b^{2}

This simplifies to the following algebraic expression:

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

Solving for the Ratio

By subtracting

a^{2}

from both sides of the equation, we isolate the relationship between the squares of the semi-axes:

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

This ratio is the key to unlocking the eccentricity of the hyperbola.

Final Calculation

The eccentricity

e

of a hyperbola is defined by the formula:

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

Substituting our derived ratio

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

into the formula, we get:

e = 1 + \frac{1}{2} = \frac{3}{2}

The eccentricity of the hyperbola is $\frac{3}{2}$ .

Common Traps & FAQs

Why is the normal equation

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

and not just the slope-point form?

You absolutely can use the slope-point form

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

, but it requires you to first find the slope of the tangent, then take its negative reciprocal to find the slope of the normal. The equation

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

is simply the derived, optimized form for the normal to a hyperbola. Using it saves time and reduces the chance of algebraic errors in the heat of an exam.

Why do we use the eccentricity formula

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

That is the fundamental definition of eccentricity for a standard hyperbola. It measures how 'stretched' the hyperbola is. Since we found the ratio

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

from the geometric properties of the normal, this formula is the direct bridge to the final answer.

Wait, why does the

y

-term vanish when we substitute

(9, 0)

Great observation! The point

(9, 0)

lies on the

x

-axis, meaning its

y

-coordinate is

0

. When you substitute

y = 0

into the equation of the normal, the term

\frac{b ^{2} y}{y _{1}}

becomes

\frac{b ^{2} ( 0 )}{3}

, which is

0

. This is a gift from the problem setter—it simplifies the equation significantly, allowing us to isolate the relationship between

a^{2}

and

b^{2}

Animated Solution for Mathematics - Conic Sections: 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

Select Answer:

Visualized Solution

Visualizing the Hyperbola and Point $P$

Point $P (6, 3)$ on the Hyperbola

Equation of the Normal

Substituting Point $P (6, 3)$

Using the $x$ -axis Intersection

Simplifying the Equation

Finding the Relation Between $a^{2}$ and $b^{2}$

Formula for Eccentricity $e$

Calculating Eccentricity $e$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal Line Equation

The Intersection Constraint

Solving for the Ratio

Final Calculation

Similar Questions

Let $P (3, 3)$ be a point on the hyperbola, $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If the normal to it at $P$ intersects the x-axis at $(9, 0)$ and $e$ is its eccentricity, then the ordered pair $(a^{2}, e^{2})$ is equal to :

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 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-

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 $P (a sec θ, b tan θ)$ and $Q (a sec ϕ, b tan ϕ)$ , where $θ + ϕ = π /2$ , be two points on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$ , then $k$ is equal to

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:

If a hyperbola passes through the point $P (10, 16)$ and it has vertices at $(\pm 6, 0)$ , then the equation of the normal at $P$ is:

Select Answer:

Visualized Solution

Visualizing the Hyperbola and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(6,3) on the Hyperbola

Equation of the Normal

Substituting Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(6,3)

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

Simplifying the Equation

Finding the Relation Between .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

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

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

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal Line Equation

The Intersection Constraint

Solving for the Ratio

Final Calculation

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; } 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:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a hyperbola passes through the point P(10,16) and it has vertices at (±6,0), then the equation of the normal at P is:

Visualizing the Hyperbola and Point $P$

Point $P (6, 3)$ on the Hyperbola

Substituting Point $P (6, 3)$

Using the $x$ -axis Intersection

Finding the Relation Between $a^{2}$ and $b^{2}$

Formula for Eccentricity $e$

Calculating Eccentricity $e$

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:

If a hyperbola passes through the point $P (10, 16)$ and it has vertices at $(\pm 6, 0)$ , then the equation of the normal at $P$ is: