The tangent to `y^(2)=ax` make angles `theta_(1)andtheta_(2)` with the x-axis. If `costheta_(1)costheta_(2)=lamda`, then the locus of their point of intersection is

Options

A. `x^(2)=lamda^(2)[(x-a)^(2)+4y^(2)]`
B. `x^(2)=lamda^(2)[(+-a)^(2)+y^(2)]`
C. `x^(2)=lamda^(2)[(x-a)^(2)+y^(2)]`
D. `4x^(2)=lamda^(2)[(x-a)^(2)+y^(2)]`

Answer

C

Explanation

(3) Let the tangents at `P(at_(1)^(2),2at_(1))andQ(at_(2)^(2),2at_(2))` make angles `theta_(1)andtheta_(2)` with x-axis.
Then `"tan"theta_(1)/(t_(1))andtantheta_(2)=(1)/(t_(2))`.
Given that
`costheta_(1)costheta_(2)=lamda`
`rArrsec^(2)theta_(1)sec^(2)theta_(2)=(1)/(lamda^(2))`
`rArr(1+tan^(2)theta_(1))(1+tan^(2)theta_(2))=(1)/(lamda^(2))`
`rArr(1+(1)/(t_(1)^(2)))(1+(1)/(t_(2)^(2)))=(1)/(lamda^(2))`
`rArrlamda^(2)[1+(t_(1)+t_(2))^(2)-2t_(1)t_(2)+t_(1)^(2)t_(2)^(2)]=t_(1)^(2)t_(2)^(2)`
Now, point of intersection of tangents is
`(at_(1)t_(2),a(t_(1)+t_(2)))-=(h,k)`
`rArrlamda^(2)[1+(k^(2))/(a^(2))-(2h)/(a)+(h^(2))/(a^(2))]=(h^(2))/(a^(2))`
`rArrx^(2)=lamda^(2)[(x-a)^(2)+y^(2)]`