IPMAT Indore 2021 (SA) - If one of the lines given by the equation 2x^2 + axy + 3y^2 = 0 coincides with one of those given by 2x^2 + bxy - 3y^2 = 0 and the other lines represented by them are perpendicular then a^2+b^2 = | PYQs + Solutions

IPMAT Indore 2021

Geometry

Straight Lines

Hard

If one of the lines given by the equation $2x^2 + axy + 3y^2 = 0$ coincides with one of those given by $2x^2 + bxy - 3y^2 = 0$ and the other lines represented by them are perpendicular then $a^2+b^2 =$

Entered answer:

✅ Correct Answer: 26

Alternate Solution

Let's find the value of

a^2 + b^2

when the two quadratic equations share a common line.

We have two quadratic equations:

\newline

2x^2 + axy + 3y^2 = 0

\newline

2x^2 + bxy - 3y^2 = 0

Where:

One line from the first equation coincides with one line from the second equation

The remaining lines are perpendicular to each other

For a quadratic equation

Ax^2 + Bxy + Cy^2 = 0

, if it represents two lines with slopes

m_1

and

m_2

, then:

Am_1m_2 = C

(A(m_1+m_2)) = B

For the first equation

2x^2 + axy + 3y^2 = 0

2m_1m_2 = 3

2(m_1+m_2) = a

For the second equation

2x^2 + bxy - 3y^2 = 0

2n_1n_2 = -3

2(n_1+n_2) = b

Let's say

m_1 = n_1

(one line coincides).

Then:

m_1m_2 = \dfrac{3}{2}

and

m_1n_2 = -\dfrac{3}{2}

This means

m_2 = \dfrac{3}{2m_1}

and

n_2 = -\dfrac{3}{2m_1}

For the other lines to be perpendicular:

m_2 \cdot n_2 = -1

Substituting:

\dfrac{3}{2m_1} \cdot (-\dfrac{3}{2m_1}) = -1

-\dfrac{9}{4m_1^2} = -1

\dfrac{9}{4m_1^2} = 1

m_1^2 = \dfrac{9}{4}

m_1 = \pm \dfrac{3}{2}

Calculating

a

and

b

using

m_1 = \dfrac{3}{2}

a = -2(m_1+m_2) = -2(\dfrac{3}{2}+\dfrac{3}{2 \cdot \dfrac{3}{2}}) = -2(\dfrac{3}{2}+\dfrac{3}{3}) = -2(\dfrac{3}{2}+1) = -5

b = -2(n_1+n_2) = -2(\dfrac{3}{2}-\dfrac{3}{2 \cdot \dfrac{3}{2}}) = -2(\dfrac{3}{2}-1) = -1

Therefore:

a^2 + b^2 = (-5)^2 + (-1)^2 = 25 + 1 = 26