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Animated Solution for Mathematics - Straight Lines: The pair of lines represented by $3 a x^{2} + 5 x y + (a^{2} - 2) y^{2} = 0$ are perpendicular to each other for

Q: Wait, why do we only look at the coefficients of $x^2$ and $y^2$? What about the $xy$ term?

That is a fantastic observation! In the general equation $Ax^2 + 2Hxy + By^2 = 0$, the $xy$ term (represented by $2H$) dictates the 'tilt' or the rotation of the lines. However, the condition for perpendicularity, $A + B = 0$, is derived from the fact that the sum of the slopes of the two lines is $-2H/B$ and their product is $A/B$. If the lines are perpendicular, the product of their slopes must be $-1$. When you set $m_1 m_2 = -1$, the $H$ term effectively cancels out, leaving us with the beautiful simplicity of $A + B = 0$.

Q: Is it possible for $a^2 - 2$ to be zero?

Great catch. If $a^2 - 2 = 0$, the equation would no longer be a quadratic in $y$, and one of the lines would become a vertical line ($x=0$). If one line is vertical, the other must be horizontal ($y=0$) to be perpendicular. If $y=0$ is a line, then the coefficient of $x^2$ must also be zero. Since $3a$ and $a^2-2$ cannot both be zero simultaneously, we don't have to worry about the 'degenerate' case here. Your intuition for edge cases is exactly what will get you into an IIT!

Select Answer:

Visualized Solution

Visualizing the Equation

Given equation: $3 a x^{2} + 5 x y + (a^{2} - 2) y^{2} = 0$
This is a homogeneous equation of degree $2$ in $x$ and $y$ .
It represents two straight lines passing through the origin $(0, 0)$ .

Standard Form

Standard form: $A x^{2} + 2 H x y + B y^{2} = 0$
Comparing helps identify the coefficients of $x^{2}$ and $y^{2}$ .

Perpendicularity Condition

Condition for perpendicular lines: $A + B = 0$
The sum of coefficients of $x^{2}$ and $y^{2}$ must be zero.

Identifying Coefficients

Coefficient of $x^{2}$ is $A = 3 a$
Coefficient of $y^{2}$ is $B = a^{2} - 2$

Applying the Condition

Substitute $A$ and $B$ into $A + B = 0$ :
$(3 a) + (a^{2} - 2) = 0$

Forming the Quadratic

Rearranging into standard quadratic form:
$a^{2} + 3 a - 2 = 0$

Checking the Discriminant

To find the number of real values of $a$ , check the discriminant $D$ .
The discriminant determines the nature of the roots.

Calculating $D$

$D = (3)^{2} - 4 (1) (- 2)$
$D = 9 + 8 = 17$

Interpreting $D$

Since $D = 17 > 0$ , the quadratic equation has two distinct real roots.

Final Conclusion

Therefore, the lines are perpendicular for exactly two values of $a$ .

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The Sigma Insight: Angle Between Two Lines

Solution Diagram

The Geometry of Orthogonality

A Journey into Homogeneous Equations

Imagine you are standing at the origin of a coordinate plane. You have two lines, both passing through your feet, stretching out into the infinite expanse of the

x y

-plane.

You want these lines to be perfectly perpendicular—a perfect cross, a symbol of absolute balance. We can force this condition using the elegant world of homogeneous equations of the second degree.

The Canvas

Homogeneous Equations

We start with the given equation:

3 a x^{2} + 5 x y + (a^{2} - 2) y^{2} = 0

At first glance, this might look like just another jumble of variables. However, every term here is of degree 2.

The

x^{2}

term, the

x y

term, and the

y^{2}

term all share the same weight. This is a homogeneous equation of degree 2, which is the geometric signature of a pair of straight lines passing through the origin

(0, 0)

The Condition of Perpendicularity

To understand how these lines behave, we compare our equation to the standard form:

A x^{2} + 2 H x y + B y^{2} = 0

By matching the coefficients, we identify our players:

A = 3 a

2 H = 5

, and

B = a^{2} - 2

The condition for these two lines to be perpendicular is:

A + B = 0

This occurs because if the slopes of the lines are

m_{1}

and

m_{2}

, they must satisfy

m_{1} m_{2} = - 1

. When you derive the relationship between the coefficients and the slopes, the

H

term vanishes, leaving us with the requirement that the sum of the coefficients of

x^{2}

and

y^{2}

must be zero.

The Algebraic Resolution

Now, we substitute our values into the condition

A + B = 0

(3 a) + (a^{2} - 2) = 0

Rearranging this, we find a quadratic equation in

a

a^{2} + 3 a - 2 = 0

To determine how many values of

a

satisfy this, we calculate the discriminant

D = b^{2} - 4 a c

, where

b = 3

a = 1

, and

c = - 2

D = (3)^{2} - 4 (1) (- 2) = 9 + 8 = 17

Since

D = 17

, which is strictly greater than zero, we know with absolute certainty that this quadratic equation yields two distinct, real roots.

The Final Insight

We started with a general equation, identified its geometric nature, and applied the condition for perpendicularity. Because the discriminant is positive, we conclude that there are exactly two values of $a$ for which these lines will stand at a perfect

9 0^{\circ}

angle to each other.

Common Traps & FAQs

Wait, why do we only look at the coefficients of

x^{2}

and

y^{2}

? What about the

x y

term?

That is a fantastic observation! In the general equation

A x^{2} + 2 H x y + B y^{2} = 0

, the

x y

term (represented by

2 H

) dictates the 'tilt' or the rotation of the lines. However, the condition for perpendicularity,

A + B = 0

, is derived from the fact that the sum of the slopes of the two lines is

- 2 H / B

and their product is

A / B

. If the lines are perpendicular, the product of their slopes must be

- 1

. When you set

m_{1} m_{2} = - 1

, the

H

term effectively cancels out, leaving us with the beautiful simplicity of

A + B = 0

How do I know if the lines are actually real? Could they be imaginary?

You are thinking like a true mathematician! For the lines to be real, the discriminant of the quadratic form must be non-negative. Specifically, for

A x^{2} + 2 H x y + B y^{2} = 0

, we need

H^{2} - A B \geq 0

. In this problem, we are asked for the values of

a

that make them perpendicular. Since we found

D = 17 > 0

for our quadratic in

a

, we are guaranteed two distinct real values for

a

. Checking the reality of the lines themselves is a great habit to maintain!

Why is the equation homogeneous? Does that matter?

It matters immensely! A homogeneous equation of degree 2 means that if

(x, y)

is a solution, then

(k x, k y)

is also a solution for any scalar

k

. Geometrically, this means the solution set is a collection of lines passing through the origin. If there were a constant term or linear terms like

x

y

, the lines would be shifted away from the origin. Homogeneity is the 'origin-anchoring' property.

Can I just use the slope formula

m_{1} + m_{2}

and

m_{1} m_{2}

Absolutely! You can treat the equation as a quadratic in terms of

m = y / x

. Dividing the entire equation by

x^{2}

, you get

(a^{2} - 2) m^{2} + 5 m + 3 a = 0

. If the lines are perpendicular, the product of the roots

m_{1} m_{2}

must be

- 1

. Using the property of roots,

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

. Solving this gives

3 a = - a^{2} + 2

, which is exactly the same quadratic we derived. Both paths lead to the same truth!

Is it possible for

a^{2} - 2

to be zero?

Great catch. If

a^{2} - 2 = 0

, the equation would no longer be a quadratic in

y

, and one of the lines would become a vertical line (

x = 0

). If one line is vertical, the other must be horizontal (

y = 0

) to be perpendicular. If

y = 0

is a line, then the coefficient of

x^{2}

must also be zero. Since

3 a

and

a^{2} - 2

cannot both be zero simultaneously, we don't have to worry about the 'degenerate' case here. Your intuition for edge cases is exactly what will get you into an IIT!

Animated Solution for Mathematics - Straight Lines: The pair of lines represented by $3 a x^{2} + 5 x y + (a^{2} - 2) y^{2} = 0$ are perpendicular to each other for

Select Answer:

Visualized Solution

Visualizing the Equation

Standard Form

Perpendicularity Condition

Identifying Coefficients

Applying the Condition

Forming the Quadratic

Checking the Discriminant

Calculating $D$

Interpreting $D$

Final Conclusion

The Sigma Insight: Angle Between Two Lines

The Geometry of Orthogonality

The Canvas

The Condition of Perpendicularity

The Algebraic Resolution

The Final Insight

Similar Questions

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The pair of lines represented by 3ax2+5xy+(a2−2)y2=0 are perpendicular to each other for

Select Answer:

Visualized Solution

Visualizing the Equation

Standard Form

Perpendicularity Condition

Identifying Coefficients

Applying the Condition

Forming the Quadratic

Checking the Discriminant

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

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

Final Conclusion

The Sigma Insight: Angle Between Two Lines

The Geometry of Orthogonality

The Canvas

The Condition of Perpendicularity

The Algebraic Resolution

The Final Insight

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Animated Solution for Mathematics - Straight Lines: The pair of lines represented by $3 a x^{2} + 5 x y + (a^{2} - 2) y^{2} = 0$ are perpendicular to each other for

Calculating $D$

Interpreting $D$

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