If the tangents and normals at the extremities of a focal chord of a parabola intersect at `(x_1,y_1)` and `(x_2,y_2),` respectively, then `x_1=y^2` (b) `x_1=y_1` `y_1=y_2` (d) `x_2=y_1`

Options

A. `x_(1)=y_(2)`
B. `x_(1)=y_(1)`
C. `y_(1)=y_(2)`
D. `x_(2)=y_(1)`

Answer

C

Explanation

(3) Let P `(at_(1)^(2),2at_(1))andQ(at_(2)^(2),at_(2))` be the extremities of a focal chord of the parabola `y^(2)=4ax`. The tangents at P and Q intersect
`at(at_(1)t_(2),a(t_(1)+t_(2)))`. Therefore,
`x_(1)=at_(1)t_(2)andy_(1)=a(t_(1)+t_(2))`
`orx_(1)=-aandy_(1)=a(t_(1)+t_(2))`
`[becausePQ" is a focal chord ",t_(1)t_(2)=-1]`
The normals at P and Q intersect at
`(2a+a(t_(1)^(2)+t_(2)^(2)+t_(1)t_(2)),-at_(1)t_(2)(t_(1)+t_(2)))`
`:." "x_(2)=2a+a(t_(1)^(2)+t_(2)^(2)+t_(1)t_(2))`
`andy_(2)=-at_(1)t_(2)(t_(1)+t_(2))`
`orx_(2)=2a+a(t_(1)^(2)+t_(2)^(2)-1)=a+a(t_(1)^(2)+t_(2)^(2))`
`andy^(2)=a(t_(1)+t_(2))`
clearly, `y_(1)=y_(2)`.