Question

If $(x_1,y_1)(x_2,y_2)$ are the extremities of a focal chord of the parabola $y^2=16x$ then $4x_1{x}_2+y_1{y}_2=$

• A. -48
• B. 0
• C. -64
• D. 16

Answer 1

The correct option is B

0

$y^2=16x$

a=4
$(x_1,y_1)=(a{t}_1^2,2a{t}_1)$
$(x_2,y_2)=(a{t}_2^2,2a{t}_2)$
$4x_1{x}_2+y_1{y}_2=4a^2{t}_1^2{t}_2^2+4a^2{t}_1{t}_2)$
$4\times 16\times (-1{)}^2+4\times 16\times -1=64-64=0$