JEE Main 2018 (15 April Evening)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: 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 :-

Select Answer:

Visualized Solution

Standard Form of Hyperbola

Given equation: $4 x^{2} - 9 y^{2} = 36$
Divide by $36$ to get standard form: $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$

Extracting Parameters

Comparing with $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
$a^{2} = 9 ⟹ a = 3$
$b^{2} = 4 ⟹ b = 2$

Equation of the Normal

General equation of normal at $(a sec θ, b tan θ)$ :
$\frac{a x}{s e c θ} + \frac{b y}{t a n θ} = a^{2} + b^{2}$

Substituting Parameters

Substitute $a = 3$ and $b = 2$ :
$\frac{3 x}{s e c θ} + \frac{2 y}{t a n θ} = 9 + 4$
Simplified: $3 x cos θ + 2 y cot θ = 13$

Finding Intercept A

The normal meets the x-axis at $A$ .
Set $y = 0$ in the normal equation.
$3 x cos θ = 13 ⟹ x_{A} = \frac{13}{3 c o s θ}$

Finding Intercept B

The normal meets the y-axis at $B$ .
Set $x = 0$ in the normal equation.
$2 y cot θ = 13 ⟹ y_{B} = \frac{13}{2 c o t θ} = \frac{13 t a n θ}{2}$

Geometry of Parallelogram OABP

$O A B P$ is a parallelogram with $O$ at the origin.
Since axes $O A$ and $O B$ are perpendicular, $O A B P$ is a rectangle.
Let the coordinates of $P$ be $(h, k)$ .

Coordinates of P

For rectangle $O A B P$ , $P$ shares its x-coordinate with $A$ and y-coordinate with $B$ .
$h = x_{A} = \frac{13}{3 c o s θ}$
$k = y_{B} = \frac{13 t a n θ}{2}$

Isolating Trigonometric Terms

From $h = \frac{13}{3 c o s θ}$ , we get $sec θ = \frac{3 h}{13}$
From $k = \frac{13 t a n θ}{2}$ , we get $tan θ = \frac{2 k}{13}$

Eliminating the Parameter $θ$

Use the fundamental identity: $sec^{2} θ - tan^{2} θ = 1$
Substitute the expressions for $sec θ$ and $tan θ$ :
$(\frac{3 h}{13})^{2} - (\frac{2 k}{13})^{2} = 1$

Final Locus Equation

Expand the squares: $\frac{9 h ^{2}}{169} - \frac{4 k ^{2}}{169} = 1$
Multiply by $169$ : $9 h^{2} - 4 k^{2} = 169$
Replace $(h, k)$ with $(x, y)$ : $9 x^{2} - 4 y^{2} = 169$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given equation of the hyperbola is

4 x^{2} - 9 y^{2} = 36

. To standardize this, we divide the entire equation by

36

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

Comparing this to the standard form

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

, we identify the parameters as

a^{2} = 9

(so

a = 3

) and

b^{2} = 4

(so

b = 2

). These values define the geometry of our hyperbola.

The Normal's Path

The equation of the normal to the hyperbola at a parametric point

(a sec θ, b tan θ)

is given by:

\frac{a x}{sec θ} + \frac{b y}{tan θ} = a^{2} + b^{2}

Substituting

a = 3

and

b = 2

into this formula, we obtain:

\frac{3 x}{sec θ} + \frac{2 y}{tan θ} = 9 + 4

Using the trigonometric identities

\frac{1}{s e c θ} = cos θ

and

\frac{1}{t a n θ} = cot θ

, the equation simplifies to:

3 x cos θ + 2 y cot θ = 13

The Geometry of the Parallelogram

The normal intersects the axes at points

A

and

B

. To find the coordinates of

A

, we set

y = 0

3 x cos θ = 13 ⟹ x_{A} = \frac{13}{3 cos θ}

To find the coordinates of

B

, we set

x = 0

2 y cot θ = 13 ⟹ y_{B} = \frac{13}{2 cot θ} = \frac{13 tan θ}{2}

Since the axes are perpendicular, the parallelogram

O A B P

is a rectangle. Therefore, the coordinates of point

P (h, k)

are:

(h, k) = (\frac{13}{3 cos θ}, \frac{13 tan θ}{2})

Final Calculation

To find the locus, we eliminate the parameter

θ

using the identity

sec^{2} θ - tan^{2} θ = 1

. From our coordinates, we have:

sec θ = \frac{3 h}{13} and tan θ = \frac{2 k}{13}

Substituting these into the identity yields:

(\frac{3 h}{13})^{2} - (\frac{2 k}{13})^{2} = 1

Expanding and simplifying, we get:

\frac{9 h ^{2}}{169} - \frac{4 k ^{2}}{169} = 1 ⟹ 9 h^{2} - 4 k^{2} = 169

Replacing

(h, k)

with

(x, y)

, the final locus of point

P

is:

$9 x^{2} - 4 y^{2} = 169$

Common Traps & FAQs

Wait, why is OABP a rectangle? I thought it was just a parallelogram.

That is a great observation! A parallelogram is defined by having opposite sides parallel. However, in this problem, the vertices A and B lie on the coordinate axes. Since the x-axis and y-axis are perpendicular to each other, the angle at the origin O is

9 0^{\circ}

. A parallelogram with at least one right angle is, by definition, a rectangle. This simplifies our life immensely because the coordinates of P are simply the x-coordinate of A and the y-coordinate of B.

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 can! However, using the slope form often leads to messy algebraic expressions involving square roots and quadratic equations. The parametric form, using

θ

, is much more elegant for locus problems because it allows us to use the fundamental trigonometric identity

sec^{2} θ - tan^{2} θ = 1

to eliminate the parameter cleanly. In JEE Advanced, choosing the right tool is half the battle.

Why did we get

9 x^{2} - 4 y^{2} = 169

instead of the original hyperbola

4 x^{2} - 9 y^{2} = 36

Think of the locus as the 'path' traced by point P as the normal line moves along the hyperbola. The original hyperbola is the set of points where the normal lines originate. The locus of P is a different curve entirely, generated by the geometric constraints of the intercepts. It is perfectly normal for the resulting locus to have a different equation!

Animated Solution for Mathematics - Conic Sections: 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 :-

Select Answer:

Visualized Solution

Standard Form of Hyperbola

Extracting Parameters

Equation of the Normal

Substituting Parameters

Finding Intercept A

Finding Intercept B

Geometry of Parallelogram OABP

Coordinates of P

Isolating Trigonometric Terms

Eliminating the Parameter $θ$

Final Locus Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal's Path

The Geometry of the Parallelogram

Final Calculation

Similar Questions

If the tangents drawn to the hyperbola $4 y^{2} = x^{2} + 1$ intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :

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:

Comprehension Passage

The coordinates of $A$ and $B$ are

The orthocenter of the triangle $P A B$ is

The equation of the locus of the point whose distances from the point $P$ and the line $A B$ are equal, is

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

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

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The locus of a point $P (α, β)$ moving under the condition that the line $y = α x + β$ is a tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is

Select Answer:

Visualized Solution

Standard Form of Hyperbola

Extracting Parameters

Equation of the Normal

Substituting Parameters

Finding Intercept A

Finding Intercept B

Geometry of Parallelogram OABP

Coordinates of P

Isolating Trigonometric Terms

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

Final Locus Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal's Path

The Geometry of the Parallelogram

Final Calculation

Similar Questions

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.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:

Comprehension Passage

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tangents are drawn to the hyperbola 9x2​−4y2​=1, parallel to the straight line 2x−y=1. The points of contact of the tangents on the hyperbola are

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

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The locus of a point $P (α, β)$ moving under the condition that the line $y = α x + β$ is a tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is