Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1760 Upvotes
Verified Answer
The correct answer is:
$\left(\frac{9}{2}, 0\right)$
$$
\begin{array}{l}
(2 x-4)^{2}=4 x \\
(x-2)^{2}=x \\
x^{2}-5 x+4=0 \\
x=1,4
\end{array}
$$
$$
\begin{array}{l}
\mathrm{C}(1,-2) \\
\mathrm{B}(4,4) \quad \because \mathrm{AB}=\mathrm{AC} \\
\sqrt{(\alpha-4)^{2}+16}=\sqrt{(\alpha-1)^{2}+4}
\end{array}
$$
On solving, we get $\alpha=\frac{9}{2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.