JEE Main 2020 - 4 Sep (Morning)

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Animated Solution for Mathematics - Conic Sections: 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 :

Q: Wait, why is the normal equation $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$? I thought it was different.

That is a common point of confusion! The equation $\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$ is the equation of the *tangent* at $(x_1, y_1)$. The normal is the line perpendicular to the tangent at that point. The formula $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$ is the derived standard form for the normal to a hyperbola. It is a powerful tool, so keep it in your memory bank!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the hyperbola

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

and the point

P (3, 3)

. The normal at

P

is a line that is perpendicular to the tangent at

P

. The crucial geometric constraint is that this normal line passes through the point

(9, 0)

on the x-axis. This means the coordinates

(9, 0)

must satisfy the equation of the normal.

Hint 2

Hint 2: The Arsenal. Recall the standard equation of the normal to the hyperbola

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

at the point

(x_{1}, y_{1})

, which is

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

. You also need the definition of eccentricity

e

for a hyperbola, which is

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

Hint 3

Hint 3: The Execution. Substitute

x_{1} = 3

and

y_{1} = 3

into the normal equation. Then, substitute

x = 9

and

y = 0

into that equation to find a relationship between

a^{2}

and

b^{2}

. Use this relationship in the original hyperbola equation

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

to solve for

a^{2}

Visualized Solution

Visualizing the Hyperbola and Point $P$

Standard Hyperbola: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Point $P (3, 3)$ lies on the curve.

Substituting Point $P (3, 3)$

Substitute $x = 3, y = 3$ :
$\frac{3 ^{2}}{a ^{2}} - \frac{3 ^{2}}{b ^{2}} = 1$
$\frac{9}{a ^{2}} - \frac{9}{b ^{2}} = 1 \dots (1)$

Equation of the Normal

The normal at $(x_{1}, y_{1})$ to $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is:
$\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}$

Normal at $P (3, 3)$

Substitute $x_{1} = 3, y_{1} = 3$ :
$\frac{a ^{2} x}{3} + \frac{b ^{2} y}{3} = a^{2} + b^{2}$

Intersection with X-axis

The normal intersects the x-axis at $(9, 0)$ .
Substitute $x = 9, y = 0$ into the normal equation.

Simplifying the Normal Equation

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

Solving for $a^{2}$

Substitute $b^{2} = 2 a^{2}$ into equation (1):
$\frac{9}{a ^{2}} - \frac{9}{2 a ^{2}} = 1$
$\frac{18 - 9}{2 a ^{2}} = 1$

Calculating $a^{2}$

$\frac{9}{2 a ^{2}} = 1$
$2 a^{2} = 9$
$a^{2} = \frac{9}{2}$

Eccentricity Formula

Eccentricity $e$ of a hyperbola:
$e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$

Calculating $e^{2}$

We know $b^{2} = 2 a^{2}$ , so $\frac{b ^{2}}{a ^{2}} = 2$
$e^{2} = 1 + 2$
$e^{2} = 3$

Final Answer

The ordered pair $(a^{2}, e^{2})$ is $(\frac{9}{2}, 3)$ .
Option (2) is correct.

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler in the world of coordinate geometry! Today, we are going to unravel the secrets of a hyperbola defined by the equation:

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

We are given a specific point

P (3, 3)

resting on its branch. This point serves as the key to unlocking the parameters

a^{2}

and

b^{2}

The Point of Contact

Since

P (3, 3)

lies on the hyperbola, it must satisfy the hyperbola's equation. By substituting

x = 3

and

y = 3

, we obtain:

\frac{3 ^{2}}{a ^{2}} - \frac{3 ^{2}}{b ^{2}} = 1 \Rightarrow \frac{9}{a ^{2}} - \frac{9}{b ^{2}} = 1

Let us hold onto this as our first equation. It is a simple statement of existence, but it is the foundation of everything that follows.

The Geometry of the Normal

Now, let us consider the normal at

P

. The normal is the line perpendicular to the tangent at

P

. For a hyperbola, the equation of the normal at any point

(x_{1}, y_{1})

is given by the formula:

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

By substituting our point

P (3, 3)

, the equation becomes:

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

The Algebraic Bridge

The problem states that this normal line intersects the x-axis at

(9, 0)

. If the line passes through

(9, 0)

, then these coordinates must satisfy the equation of the normal.

Substituting

x = 9

and

y = 0

into our normal equation, we get:

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

The

y

term vanishes, leaving us with

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

. With a quick rearrangement, we find the hidden symmetry:

2 a^{2} = b^{2}

The Final Synthesis

Now, we return to our first equation:

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

. Substituting

b^{2} = 2 a^{2}

, we get:

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

Combining the fractions, we have

\frac{18 - 9}{2 a ^{2}} = 1

, which simplifies to

\frac{9}{2 a ^{2}} = 1

. Thus,

2 a^{2} = 9

, or $a^{2} = \frac{9}{2}$ .

Finally, we calculate the eccentricity

e

using the formula

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

. Since we know

b^{2} = 2 a^{2}

, the ratio

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

is simply

2

Therefore,

e^{2} = 1 + 2 = 3

. Our final result for the ordered pair

(a^{2}, e^{2})

is $(\frac{9}{2}, 3)$ .

Common Traps & FAQs

Wait, why is the normal equation

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

? I thought it was different.

That is a common point of confusion! The equation

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

is the equation of the *tangent* at

(x_{1}, y_{1})

. The normal is the line perpendicular to the tangent at that point. The formula

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

is the derived standard form for the normal to a hyperbola. It is a powerful tool, so keep it in your memory bank!

Why can't we just use the slope form of the normal,

y = m x \pm \frac{m ( a ^{2} + b ^{2} )}{a ^{2} - b ^{2} m ^{2}}

You absolutely could! However, the slope form is often more algebraically intensive because you would first need to find the slope of the tangent at

(3, 3)

, then find the negative reciprocal for the normal's slope, and finally use the point-slope form. Using the point form of the normal is much faster and less prone to calculation errors in this specific problem.

Why did the

y

term vanish when we substituted

(9, 0)

Great observation! The normal line intersects the x-axis at

(9, 0)

. By definition, any point on the x-axis has a y-coordinate of

0

. When we substitute

y = 0

into the equation of the line, the entire term containing

y

becomes zero, which simplifies the equation significantly and allows us to isolate the relationship between

a^{2}

and

b^{2}

Animated Solution for Mathematics - Conic Sections: 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 :

Select Answer:

Visualized Solution

Visualizing the Hyperbola and Point $P$

Substituting Point $P (3, 3)$

Equation of the Normal

Normal at $P (3, 3)$

Intersection with X-axis

Simplifying the Normal Equation

Solving for $a^{2}$

Calculating $a^{2}$

Eccentricity Formula

Calculating $e^{2}$

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Point of Contact

The Geometry of the Normal

The Algebraic Bridge

The Final Synthesis

Similar Questions

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

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 :

The normal to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{9} = 1$ at the point $(8, 33)$ on it passes through the point :

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:

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 ______.

A normal to the hyperbola, $4 x^{2} - 9 y^{2} = 36$ meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the locus of P is :-

Let the tangent to the parabola $y^{2} = 12 x$ at the point $(3, α)$ be perpendicular to the line $2 x + 2 y = 3$ . Then the square of distance of the point $(6, - 4)$ from the normal to the hyperbola $α^{2} x^{2} - 9 y^{2} = 9 α^{2}$ at its point $(α - 1, α + 2)$ is equal to .............

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

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

Equation of the Normal

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

Intersection with X-axis

Simplifying the Normal Equation

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

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

Eccentricity Formula

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e2

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Point of Contact

The Geometry of the Normal

The Algebraic Bridge

The Final Synthesis

Similar Questions

.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; } The normal to the hyperbola a2x2​−9y2​=1 at the point (8,33​) on it passes through the point :

.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:

Visualizing the Hyperbola and Point $P$

Substituting Point $P (3, 3)$

Normal at $P (3, 3)$

Solving for $a^{2}$

Calculating $a^{2}$

Calculating $e^{2}$

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

The normal to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{9} = 1$ at the point $(8, 33)$ on it passes through the point :

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: