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

• A

  $\left(\frac{5}{2}, 0\right)$
• B

  $\left(\frac{7}{2}, 0\right)$
• Correct answer
  C

  $\left(\frac{9}{2}, 0\right)$
• D

  $\left(\frac{11}{2}, 0\right)$

Solution

$$ \begin{array}{l} (2 x-4)^{2}=4 x \\ (x-2)^{2}=x \\ x^{2}-5 x+4=0 \\ x=1,4 \end{array} $$ $$ \begin{array}{l} \mathrm{C}(1,-2) \\ \mathrm{B}(4,4) \quad \because \mathrm{AB}=\mathrm{AC} \\ \sqrt{(\alpha-4)^{2}+16}=\sqrt{(\alpha-1)^{2}+4} \end{array} $$ On solving, we get $\alpha=\frac{9}{2}$

Hence, the correct answer is option: 'C'.