The locus of a point $P(\alpha, \beta)$ moving under the condition that the line $y = \alpha x + \beta$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is Options: (A) an ellipse (B) a circle (C) a parabola (D) a hyperbola

### Analyzing the Setup We are tasked with finding the locus of a point $P(\alpha, \beta)$ such that the line $y = \alpha x + \beta$ is a tangent to the hyperbola defined by: $$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $$ The hyperbola is centered at the origin $(0, 0)$ with its branches opening along the $x$-axis. The equation serves as the blueprint for the curve's geometry. ### The Condition of Tangency For a general line $y = mx + c$ to be tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, it must satisfy the specific condition of tangency: $$ c^2 = a^2m^2 - b^2 $$ This formula acts as the mathematical bridge between the line's parameters and the hyperbola's dimensions, ensuring the line "kisses" the curve at exactly one point. ### Mapping and Substitution We compare our given line $y = \alpha x + \beta$ to the general form $y = mx + c$. By direct comparison, we identify the slope $m = \alpha$ and the $y$-intercept $c = \beta$. Substituting these values into the tangency condition, we obtain: $$ \beta^2 = a^2\alpha^2 - b^2 $$ This equation represents the fundamental relationship that the coordinates $(\alpha, \beta)$ must satisfy for the line to remain tangent to the hyperbola. ### The Locus Revealed To determine the locus of the point $P(\alpha, \beta)$, we replace the parameters $\alpha$ and $\beta$ with the general coordinate variables $x$ and $y$. This yields the equation: $$ y^2 = a^2x^2 - b^2 $$ Rearranging the terms to isolate the constant, we get: $$ a^2x^2 - y^2 = b^2 $$ Dividing both sides by $b^2$, we arrive at the final form: $$ \frac{x^2}{(b/a)^2} - \frac{y^2}{b^2} = 1 $$ The locus of the point $P$ is itself a **hyperbola**. We have successfully navigated the algebraic constraints to reveal the underlying geometric truth.

JEE Main 2005

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the line

y = α x + β

as a tangent to the hyperbola. A tangent is not just any line; it is a line that intersects the curve at exactly one point. Think about what geometric condition must be satisfied for a line to be a tangent to a conic section.

Hint 2

Hint 2 (The Arsenal): Recall the standard condition for a line

y = m x + c

to be a tangent to the hyperbola

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

. This condition is

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

. This is the bridge between the algebra of the line and the geometry of the hyperbola.

Hint 3

Hint 3 (The Execution): Map the parameters of your given line

y = α x + β

to the general form

y = m x + c

. You will find

m = α

and

c = β

. Substitute these into the tangency condition

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

and replace

(α, β)

with

(x, y)

to derive the equation of the locus.

Visualized Solution

The Given Hyperbola

We are given the standard hyperbola: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Let's visualize this on the coordinate plane.

The Tangent Line

A line is given by the equation: $y = α x + β$
This line is a tangent to our hyperbola.

Condition of Tangency

For a general line $y = m x + c$ to be tangent to $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ :
The standard condition is: $c^{2} = a^{2} m^{2} - b^{2}$

Mapping the Variables

Compare our line $y = α x + β$ with $y = m x + c$ .
Slope: $m = α$
Y-intercept: $c = β$

Substituting the Values

Substitute $m = α$ and $c = β$ into the tangency condition.
$c^{2} = a^{2} m^{2} - b^{2} ⟹ β^{2} = a^{2} α^{2} - b^{2}$

Rearranging for the Locus

We need the locus of the point $P (α, β)$ .
Rearrange the equation: $a^{2} α^{2} - β^{2} = b^{2}$

Identifying the Locus

To find the general equation of the locus, replace $(α, β)$ with $(x, y)$ .
$a^{2} x^{2} - y^{2} = b^{2}$
Dividing by $b^{2}$ : $\frac{x ^{2}}{( b / a ) ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
This represents a Hyperbola.

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are tasked with finding the locus of a point

P (α, β)

such that the line

y = α x + β

is a tangent to the hyperbola defined by:

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

The hyperbola is centered at the origin

(0, 0)

with its branches opening along the

x

-axis. The equation serves as the blueprint for the curve's geometry.

The Condition of Tangency

For a general 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 specific condition of tangency:

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

This formula acts as the mathematical bridge between the line's parameters and the hyperbola's dimensions, ensuring the line "kisses" the curve at exactly one point.

Mapping and Substitution

We compare our given line

y = α x + β

to the general form

y = m x + c

. By direct comparison, we identify the slope

m = α

and the

y

-intercept

c = β

Substituting these values into the tangency condition, we obtain:

β^{2} = a^{2} α^{2} - b^{2}

This equation represents the fundamental relationship that the coordinates

(α, β)

must satisfy for the line to remain tangent to the hyperbola.

The Locus Revealed

To determine the locus of the point

P (α, β)

, we replace the parameters

α

and

β

with the general coordinate variables

x

and

y

. This yields the equation:

y^{2} = a^{2} x^{2} - b^{2}

Rearranging the terms to isolate the constant, we get:

a^{2} x^{2} - y^{2} = b^{2}

Dividing both sides by

b^{2}

, we arrive at the final form:

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

The locus of the point

P

is itself a hyperbola. We have successfully navigated the algebraic constraints to reveal the underlying geometric truth.

Common Traps & FAQs

Wait, why is the condition

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

and not

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

That is a fantastic question! It is easy to mix up the conditions for different conics. For an ellipse

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

, the condition is

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

. However, for a hyperbola

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

, the negative sign in the hyperbola equation carries over to the tangency condition, resulting in

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

. Always remember: the sign in the condition matches the sign in the conic equation!

Can we just use the discriminant method (

D = 0

) instead of the formula?

Absolutely! If you ever forget the formula, you can substitute

y = α x + β

into the hyperbola equation

\frac{x ^{2}}{a ^{2}} - \frac{( α x + β ) ^{2}}{b ^{2}} = 1

. This will give you a quadratic equation in

x

. Since the line is a tangent, it touches the curve at only one point, meaning the discriminant of this quadratic must be zero. It is a longer path, but it is a foolproof way to derive the condition yourself.

What if the line is vertical? Does the slope

m

exist?

Great catch! A vertical line has an undefined slope, so it cannot be represented by

y = m x + c

. In this specific problem, we are given the form

y = α x + β

, which explicitly assumes a non-vertical line. Therefore, we only need to consider the case where the slope is defined. You are thinking like a true mathematician by checking the edge cases!

Animated Solution for Mathematics - Conic Sections: 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

The Given Hyperbola

The Tangent Line

Condition of Tangency

Mapping the Variables

Substituting the Values

Rearranging for the Locus

Identifying the Locus

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Condition of Tangency

Mapping and Substitution

The Locus Revealed

Similar Questions

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

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 :

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 tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

If two tangents drawn from a point $P$ to the parabola $y^{2} = 16 (x - 3)$ are at right angles, then the locus of point $P$ is :

If two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

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 :-

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

Select Answer:

Visualized Solution

The Given Hyperbola

The Tangent Line

Condition of Tangency

Mapping the Variables

Substituting the Values

Rearranging for the Locus

Identifying the Locus

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Condition of Tangency

Mapping and Substitution

The Locus Revealed

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between a pair of tangents drawn from a point P to the parabola y2=4ax is 45∘. Show that the locus of the point P is a hyperbola.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn to the hyperbola 4y2=x2+1 intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB 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 tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two tangents drawn from a point P to the parabola y2=16(x−3) are at right angles, then the locus of point P is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two tangents drawn from a point P to the parabola y2=4x are at right angles, then the locus of P is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If tangents are drawn to the ellipse x2+2y2=2, then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

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 :

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 tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

If two tangents drawn from a point $P$ to the parabola $y^{2} = 16 (x - 3)$ are at right angles, then the locus of point $P$ is :

If two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is