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If y+a=m(1)(x+3a),y+a=m(2)(x+3a) are two...

If `y+a=m_(1)(x+3a),y+a=m_(2)(x+3a)` are two tangents to the parabola `y^(2)=4ax`, then

A

`m_(1)+m_(2)+m_(1)m_(2)=0`

B

`m_(1)+m_(2)-m_(1)m_(2)=2/3`

C

`m_(1)+m_(2)+2m_(1)m_(2)=1/3`

D

`m_(1)+m_(2)-2m_(1)m_(2)=1`

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
Let equation of tangent is `y=mx+a/m`
`implies-a=-3am+a/m`
`implies3m^(2)-m-1=0`
`m_(1)+m_(2)=1/3,m_(1),m_(2)=-1/3`
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