The equation of the perpendiculars drawn from the origin to the lines represented by the equation `2x^2-10xy+12y^2+5x-16y-3=0`, is

Options

A. `6x^2+5xy+y^2=0`
B. `6y^2+5xy+x^2=0`
C. `6x^2-5xy+y^2=0`
D. None of these

Answer

A

Explanation

We have , `2x^2-10xy+12y^2+5x-16y-3=0`......(i)
Therefore, `m_1+m_2=(10)/(12)=(5)/(6)`.....(ii)
`m_1.m_2=(2)/(12)=(1)/(6)`....(iii)
Now , `m_1-m_2=sqrt(m_1+m_2-4m_1m_2)=sqrt(((5)/(6))^2-4xx(1)/(6))`
`rArr m_1-m_2=sqrt((25)/(36)-(4)/(6))=sqrt((25-24)/(36))=(1)/(6)`.......(iv)
On solving Eqs. (ii) and (iv) , we get `m_1=(1)/(2),m_2=(1)/(3)`
`therefore` Slope of perpendicular will be - 2 and -3 , respectively.
Combined equation of perpendicular line is `(y+2x)(y+3x)=0`
`rArr y^2+3xy+2xy+6x^2=0`
`therefore 6x^2+5xy+y^2=0`