Question

If $(x_1,y_1)$ and $(x_2,y_2)$ are the extremities of a focal chord of the parabola $3y^2=4x,$ then $x_1{x}_2+y_1{y}_2$ is equal to $\frac{-p}{q}$ where $p,q$ are co-prime. Then value of $q-p$ is

• A. $2$
• B. $4$
• C. $1$
• D. $0$

Answer 1

The correct option is A $2$

If $(a{t}_1^2,2a{t}_1),(a{t}_2^2,2a{t}_2)$ are the ends of a focal chord of a parabola $y^2=4ax,$ then $t_1{t}_2=-1$
Here, $x_1=a{t}_1^2=\frac{1}{3}{t}_1^2,x_2=\frac{1}{3}{t}_2^2$
$y_1=\frac{2}{3}{t}_1,y_2=\frac{2}{3}{t}_2$
$\therefore x_1{x}_2+y_1{y}_2$
$=\frac{1}{9}(t_1{t}_2{)}^2+\frac{4}{9}{t}_1{t}_2=\frac{1}{9}-\frac{4}{9}=-\frac{1}{3}=\frac{-p}{q}\therefore q-p=2$