Question

The vertices of the base of an isosceles triangle lie on a parabola $y^2=4x$ and the base is a part of the line $y=2x-4$. If the third vertex of the triangle lies on the $x$-axis, its coordinates are

• A. $(\frac{5}{2},0)$
• B. $(\frac{7}{2},0)$
• C. $(\frac{9}{2},0)$
• D. $(\frac{11}{2},0)$

Answer 1

The correct option is C $(\frac{9}{2},0)$

Two vertices of the isosceles triangle lie on the parabola $y^2=4x$ as well as on the line $y=2x-4$

Substituting $y=2x-4$ into the parabola, we have

$(2x-4{)}^2=4x$

$\therefore 4x^2-16x+16=4x$

$\Rightarrow 4x^2-20x+16=0$

$\Rightarrow x^2-5x+4=0$

i.e. $x=1,4$

Correspondingly, coordinates of the points lying on the parabola are $(1,-2)$ and $(4,4)$

Let the third vertex on the x axis be $(h,0)$

Since its an isosceles triangle, $(h,0)$ is equidistant from $(1,-2)$ and $(4,4)$

$\therefore (h-1{)}^2+(0+2{)}^2=(h-4{)}^2+(0-4{)}^2$

$\Rightarrow h^2-2h+1+4=h^2-8h+16+16$

$\Rightarrow 6h=27$

i.e. $h=\frac{9}{2}$

The third vertex is $(\frac{9}{2},0)$