Saved Bookmarks
| 1. |
Two equal parabolas have the same vertex and their axes are at right angles. The length of the common tangent to them, isA. 3aB. `3sqrt2a`C. 6aD. 2a |
|
Answer» Correct Answer - B Let the two equal parabolas be `y^(2)=4ax` and `x^(2)=4ay.` The equation of any tangent to `y^(2)=4ax` is `y=mx+a/m" at the point"(a/m^(2),(2a)/m)` Now, `y=mx+a/mrArrx=y/m-a/m^(2)" ...(i)"` If if touches the parabola `x^(2)=4ay,` then `(-a)/m^(2)=a/(1//m)" "["Using : c"=a/m]` `(-a)/m^(2)=a/(1//m)" "["Using : c"=a/m]` `rArr" "m^(3)=-1rArr=-1` The common tangent touches `y^(2)=4ax" at "P(1/m^(2),(2a)/m)` So, the coordinates of the point of contact are `P(a, -2a)`. The common tangent (i) touches `x^(2)=4ay` at point Q whose coordinates are `((2a)/("1/m"),a/("a/m"^(2)))" "[becausex=my+a/m" touches "x^(2)=4ay" at "((2a)/m,a/m^(2))]` `=Q(2am, am^(2))=Q(-2a, a)" "[becausem=-1]` `:.` Length of the common tangent `=PQsqrt((a+2a)^(2)+(-2a-a)^(2))=3sqrt2a.` |
|