The number of values of a for which the pair of lines represented by 3ax2 + 5xy + (a2 - 2)y2 = 0, perpendicular to each other, isCorrect answer is '2'. Can you explain this answer?

Question

Answer

ines in pair Ax² + 2hxy + By² = 0 are prependicular to each other, if A + B = 0

3a + 2a² - 2 = 0

a² + 3a - 2 = 0

There exist two values of a as D > 0.

∴ a =

Answer

ines in pair Ax² + 2hxy + By² = 0 are prependicular to each other, if A + B = 0

3a + 2a² - 2 = 0

a² + 3a - 2 = 0

There exist two values of a as D > 0.

∴ a =

Answer

Perpendicular Lines:
In order to determine the number of values of 'a' for which the pair of lines represented by the given equation are perpendicular, we need to understand the conditions for two lines to be perpendicular.

Two lines are perpendicular if the product of their slopes is -1. The slope of a line is given by the coefficient of x divided by the coefficient of y in the equation of the line.

Equation Analysis:
The given equation is: 3ax^2 + 5xy + (a^2 - 2)y^2 = 0

To find the slopes of the lines represented by this equation, we need to convert it into the standard form of a line equation, which is in the form of ax + by + c = 0.

Converting the Equation:
Let's convert the given equation into the standard form by rearranging the terms:
3ax^2 + 5xy + (a^2 - 2)y^2 = 0
3ax^2 + 5xy + a^2y^2 - 2y^2 = 0
3ax^2 + 5xy + a^2y^2 = 2y^2

Now, let's divide the entire equation by y^2 to get rid of the quadratic terms:
(3ax^2 + 5xy + a^2y^2) / y^2 = 2y^2 / y^2
3ax^2 / y^2 + 5xy / y^2 + a^2 = 2

Simplifying further:
3a(x/y)^2 + 5(x/y) + a^2 = 2

Now, let's consider x/y as a new variable, say 'm':
3am^2 + 5m + a^2 - 2 = 0

Conditions for Perpendicular Lines:
For two lines to be perpendicular, the product of their slopes should be -1. Therefore, we need to find the values of 'a' for which the slopes of the lines represented by the given equation satisfy this condition.

Calculating the Slopes:
The slope of the line represented by the given equation can be found by differentiating the equation with respect to x/y and solving for m.

Differentiating the equation 3am^2 + 5m + a^2 - 2 = 0 with respect to m:
6am + 5 = 0
m = -5/6a

Product of Slopes:
Now, let's find the product of the slopes of the two lines represented by the given equation:
m1 * m2 = (-5/6a) * (-5/6a) = 25/36a^2

Condition for Perpendicular Lines:
To find the values of 'a' for which the lines are perpendicular, we need the product of their slopes to be -1.

Therefore, 25/36a^2 = -1
36a^2 = -25
a^2 = -25/36

Solution:
The equation a^2 = -25/36 has no real solutions since the square of a real number cannot be negative.

Therefore, there are no values of 'a' for which

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