Q.The area of the triangle formed by the lines $x^2-4y^2=0$ and $x=a$ is

Step 1. Factor the homogeneous pair. $x^2-4y^2=(x-2y)(x+2y)=0$, so the two lines are

Both pass through the origin $O(0,0)$.

Step 2. Find where each line meets $x=a$. On $x-2y=0$: $a-2y=0\implies y=\dfrac{a}{2}$, giving point $A\left(a,\dfrac{a}{2}\right)$. On $x+2y=0$: $a+2y=0\implies y=-\dfrac{a}{2}$, giving point $B\left(a,-\dfrac{a}{2}\right)$.

Step 3. Identify the triangle. The triangle formed by the two lines and $x=a$ has vertices $O(0,0)$, $A\left(a,\tfrac{a}{2}\right)$, $B\left(a,-\tfrac{a}{2}\right)$. Side $AB$ (both points at $x=a$) is vertical, with length

so it can be taken as the base of the triangle.

Step 4. Find the height. The height is the perpendicular (horizontal) distance from $O$ to the vertical line $x=a$, which is simply $a$. …

Factor the homogeneous pair into two lines through the origin, find the triangle's verti …

• Forgetting the $\tfrac12$ in the area formula and reporting $2a^2$ (option 1) …