Saved Bookmarks
| 1. |
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 isA. `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» Correct Answer - C (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)]` |
|