Question

Find the area of the triangle formed by the lines joining the focus of the parabola $y^2=4x$ to the points on it which have abscissa equal to $16$.

Answer 1

Given: $y^2=4x$

It is of the form $y^2=4ax$ where $a=1$

Focus$S(a,0)=(1,0)$

$SA=focaldistance=x_1+a=16+1=17$

$SC=15$

$AC=\sqrt{{SA}^2-{SC}^2}=\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8$

$AB=2AC=2\times 8=16$

Area of $△SAB=\frac{1}{2}\times AB\times SC=\frac{1}{2}\times 16\times 15=8\times 15=120sq.units$