Question

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

• A. one of the vertices of the triangle is $(0,\frac{9}{2})$
• B. one of the vertices of the triangle is $(4,4)$
• C. Area of the triangle is $15\text{sq. units}$
• D. Centroid of the triangle is $(\frac{2}{3},\frac{19}{6})$

Answer 1

Answer: (A), (B), (D)

The correct options are
A one of the vertices of the triangle is $(0,\frac{9}{2})$
B one of the vertices of the triangle is $(4,4)$
D Centroid of the triangle is $(\frac{2}{3},\frac{19}{6})$

$x^2=4y$ and $x=2(y-2)$
$\Rightarrow 4(y-2{)}^2=4y$
$\Rightarrow y=1,4$
$x=2(y-2)\Rightarrow x=-2,4$
$\therefore B(-2,1),C(4,4)$

Now, $AB=AC$
$\Rightarrow \sqrt{(0+2{)}^2+(y-1{)}^2}=\sqrt{(0-4{)}^2+(y-4{)}^2}$
$\Rightarrow 4+y^2-2y+1=16+y^2-8y+16$
$\Rightarrow y=\frac{9}{2}$
$\Rightarrow A(0,\frac{9}{2})$

Area of triangle
$=\frac{1}{2}\left|\begin{array}{ccc}0& \frac{9}{2}& 1\\ -2& 1& 1\\ 4& 4& 1\end{array}\right|=7.5\text{sq. units}$