Wait, why do we use $y = mx \pm \sqrt{a^2m^2 - b^2}$ instead of the point-slope form?

Great question! The point-slope form $y - y_1 = m(x - x_1)$ requires us to know the point of tangency $(x_1, y_1)$. However, we don't know that point yet. The slope-form equation $y = mx \pm \sqrt{a^2m^2 - b^2}$ is a powerful tool because it directly relates the slope $m$ to the hyperbola's parameters, allowing us to bypass the need for the point of tangency entirely.

JEE Main 2022 (26 June Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let a line $L_{1}$ be tangent to the hyperbola $\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1$ and let $L_{2}$ be the line passing through the origin and perpendicular to $L_{1}$ . If the locus of the point of intersection of $L_{1}$ and $L_{2}$ is $(x^{2} + y^{2})^{2} = α x^{2} + β y^{2}$ , then $α + β$ is equal to ______.

Enter Numerical Value:

Visualized Solution

The Hyperbola

Equation of the hyperbola: $\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1$
Here, $a^{2} = 16$ and $b^{2} = 4$ .

Equation of Tangent $L_{1}$

Standard equation of a tangent with slope $m$ :
$y = m x \pm a^{2} m^{2} - b^{2}$

Substituting $a^{2}$ and $b^{2}$

Substitute $a^{2} = 16$ and $b^{2} = 4$ :
$L_{1} : y = m x \pm 16 m^{2} - 4$

The Perpendicular Line $L_{2}$

Line $L_{2}$ passes through the origin $(0, 0)$ .
$L_{2}$ is perpendicular to $L_{1}$ ( $L_{2} ⊥ L_{1}$ ).

Slope of $L_{2}$

Since $L_{2} ⊥ L_{1}$ , product of their slopes is $- 1$ .
Slope of $L_{1} = m$
Slope of $L_{2} = - \frac{1}{m}$

Equation of $L_{2}$

Equation of line through origin with slope $- \frac{1}{m}$ :
$y = - \frac{1}{m} x$
Rearranging for $m$ : $m = - \frac{x}{y}$

The Point of Intersection

Let the point of intersection of $L_{1}$ and $L_{2}$ be $P (h, k)$ .
We need to find the locus of this point $P$ .

Substituting $P (h, k)$

Point $(h, k)$ lies on both $L_{1}$ and $L_{2}$ .
From $L_{2}$ , we get $m = - \frac{h}{k}$ .
Substitute $m$ and $(h, k)$ into $L_{1}$ :
$k = (- \frac{h}{k}) h \pm 16 (- \frac{h}{k})^{2} - 4$

Simplifying the Equation

$k = - \frac{h ^{2}}{k} \pm 16 \frac{h ^{2}}{k ^{2}} - 4$
Move $- \frac{h ^{2}}{k}$ to the left side:
$k + \frac{h ^{2}}{k} = \pm \frac{16 h ^{2} - 4 k ^{2}}{k ^{2}}$

Clearing the Denominator

Take LCM on the left side:
$\frac{k ^{2} + h ^{2}}{k} = \pm \frac{16 h ^{2} - 4 k ^{2}}{k}$
Cancel $k$ from both sides:
$h^{2} + k^{2} = \pm 16 h^{2} - 4 k^{2}$

Squaring Both Sides

To remove the square root, square both sides:
$(h^{2} + k^{2})^{2} = 16 h^{2} - 4 k^{2}$

Equation of the Locus

Replace $(h, k)$ with $(x, y)$ to get the general locus:
$(x^{2} + y^{2})^{2} = 16 x^{2} - 4 y^{2}$

Comparing with Given Equation

Given locus: $(x^{2} + y^{2})^{2} = α x^{2} + β y^{2}$
Our locus: $(x^{2} + y^{2})^{2} = 16 x^{2} - 4 y^{2}$
Comparing coefficients:
$α = 16$
$β = - 4$

Final Calculation

We need to find the value of $α + β$ .
$α + β = 16 + (- 4)$
$α + β = 12$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

The Geometry of Motion

Tracing the Locus

Imagine you are standing on a vast, flat plane, and before you lies a hyperbola defined by the equation:

\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1

Our goal is to understand the path traced by the intersection point of a tangent to this hyperbola and a line perpendicular to that tangent passing through the origin. This is a classic coordinate geometry problem.

Phase 1

The Tangent and the Slope

For the given hyperbola, we identify the parameters

a^{2} = 16

and

b^{2} = 4

. The equation of a tangent line

L_{1}

with slope

m

is given by:

y = m x \pm a^{2} m^{2} - b^{2}

Substituting our specific values, the equation for the tangent

L_{1}

becomes:

y = m x \pm 16 m^{2} - 4

Phase 2

The Perpendicular Constraint

Now, consider the second line

L_{2}

. It passes through the origin

(0, 0)

and is perpendicular to the tangent

L_{1}

. Since the slope of

L_{1}

m

, the slope of

L_{2}

must be

- \frac{1}{m}

The equation of

L_{2}

is therefore:

y = - \frac{1}{m} x

Rearranging this to isolate the slope, we find

m = - \frac{x}{y}

. This relationship allows us to link the slope of the tangent to the coordinates of any point on the line

L_{2}

Phase 3

The Locus Hunt

Let the point of intersection of

L_{1}

and

L_{2}

P (h, k)

. Since

P

lies on both lines, it must satisfy both equations. Substituting

m = - \frac{h}{k}

into the equation for

L_{1}

, we obtain:

k = (- \frac{h}{k}) h \pm 16 (- \frac{h}{k})^{2} - 4

This expression forces the tangent to pass through the point

P

while simultaneously satisfying the perpendicularity condition.

Phase 4

The Algebraic Dance

We simplify the equation by grouping terms:

k + \frac{h ^{2}}{k} = \pm \frac{16 h ^{2} - 4 k ^{2}}{k ^{2}}

Multiplying both sides by

k

yields

h^{2} + k^{2} = \pm 16 h^{2} - 4 k^{2}

. Squaring both sides to eliminate the radical, we get:

(h^{2} + k^{2})^{2} = 16 h^{2} - 4 k^{2}

Replacing

(h, k)

with the general coordinates

(x, y)

, the final locus is:

(x^{2} + y^{2})^{2} = 16 x^{2} - 4 y^{2}

Comparing this to the form

(x^{2} + y^{2})^{2} = α x^{2} + β y^{2}

, we identify

α = 16

and

β = - 4

. The final result is $α + β = 12$ .

Common Traps & FAQs

Wait, why do we use

y = m x \pm a^{2} m^{2} - b^{2}

instead of the point-slope form?

Great question! The point-slope form

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

requires us to know the point of tangency

(x_{1}, y_{1})

. However, we don't know that point yet. The slope-form equation

y = m x \pm a^{2} m^{2} - b^{2}

is a powerful tool because it directly relates the slope

m

to the hyperbola's parameters, allowing us to bypass the need for the point of tangency entirely.

Why does the intersection point

P (h, k)

satisfy both equations?

By definition, the intersection point of two lines lies on both lines. Therefore, the coordinates

(h, k)

must satisfy the equation of the tangent

L_{1}

and the equation of the perpendicular line

L_{2}

. This is the fundamental principle of locus problems: the point

P

is the 'bridge' that connects the two geometric conditions.

Is the locus always a simple curve? Why did we square both sides?

The locus is the path traced by the intersection point. We square both sides to eliminate the radical (the square root) in our equation. This is a standard algebraic technique to convert an implicit, radical-containing equation into a clean, polynomial form that we can easily compare with the given locus equation.

Enter Numerical Value:

Visualized Solution

The Hyperbola

Equation of Tangent $L_{1}$

Substituting $a^{2}$ and $b^{2}$

The Perpendicular Line $L_{2}$

Slope of $L_{2}$

Equation of $L_{2}$

The Point of Intersection

Substituting $P (h, k)$

Simplifying the Equation

Clearing the Denominator

Squaring Both Sides

Equation of the Locus

Comparing with Given Equation

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Motion

Phase 1

Phase 2

Phase 3

Phase 4

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 two tangents drawn from a point $(α, β)$ lying on the ellipse $25 x^{2} + 4 y^{2} = 1$ to the parabola $y^{2} = 4 x$ are such that the slope of one tangent is four times the other, then the value of $(10 α + 5)^{2} + (16 β^{2} + 50)^{2}$ equals ____.

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

Let $A (sec θ, 2 tan θ)$ and $B (sec ϕ, 2 tan ϕ)$ , where $θ + ϕ = π /2$ , be two points on the hyperbola $2 x^{2} - y^{2} = 2$ . If $(α, β)$ is the point of the intersection of the normals to the hyperbola at A and B, then $(2 β)^{2}$ is equal to .

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

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ 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

Enter Numerical Value:

Visualized Solution

The Hyperbola

Equation of Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Substituting .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

The Perpendicular Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

The Point of Intersection

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

Simplifying the Equation

Clearing the Denominator

Squaring Both Sides

Equation of the Locus

Comparing with Given Equation

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Motion

Phase 1

Phase 2

Phase 3

Phase 4

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 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; } The locus of a point P(α,β) moving under the condition that the line y=αx+β is a tangent to the hyperbola a2x2​−b2y2​=1 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the parabola y2=x at a point (α,β),(β>0) is also a tangent to the ellipse, x2+2y2=1, then α is equal to :

Equation of Tangent $L_{1}$

Substituting $a^{2}$ and $b^{2}$

The Perpendicular Line $L_{2}$

Slope of $L_{2}$

Equation of $L_{2}$

Substituting $P (h, k)$

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 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

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

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :