The point of lines represented by  3ax^{2}+5xy+(a^{2}-2)y^{2}=0   and \perp  to each other for

  • Option 1)

    two values of a   

  • Option 2)

    \forall \; \; a

  • Option 3)

    for one value of a

  • Option 4)

    for no values of a

 

Answers (1)

As we learnt in 

General equation of a conic -

ax^{2}+2hxy+by^{2}+2gx+2fy+c= 0

- wherein

a,b,c, f,g,h  are constants

 

 3ax^{2}+5xy+\left ( a^{2}-2 \right )y^{2}=0

For perpendicular lines, sum of coefficients of x and y=0

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

So, there are two real values of a.


Option 1)

two values of a   

This option is correct

Option 2)

\forall \; \; a

This option is incorrect

Option 3)

for one value of a

This option is incorrect

Option 4)

for no values of a

This option is incorrect

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