Question

If the parabola $x^2=ay$ makes an intercept of length $\sqrt{40}$ unit on the line $y-2x=1$ then a is equal to

• A. $1$
• B. $-2$
• C. $-1$
• D. $2$

Answer 1

Answer: (A), (B)

Solving $x^2=ay$ with $y-2x=1$,
$x^2=a\left(1+2x\right)\Rightarrow x^2-2ax-a=0$
Let $x_1$ & $x_2$ are the roots
so, ${\left(x_1-x_2\right)}^2={\left(2a\right)}^2-4\left(-a\right)=4a\left(a+1\right)$
also, ${\left(y_1-y_2\right)}^2=\left(\left(2x_1+1\right)-{\left(2x_2+1\right)}^2=4{\left(x_1-x_2\right)}^2=16a\left(a+1\right)\right)$
now $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}=\sqrt{4a\left(a+1\right)+16a\left(a+1\right)}=\sqrt{40}$
$\Rightarrow 20a\left(a+1\right)=40\Rightarrow a^2+a-2=0\Rightarrow a^2+a-2=0\Rightarrow a=-2,1$