Question

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

• A. $1$
• B. $-2$
• C. $3$
• D. $2$

Answer 1

The correct option is A $1$

$y=1+2x$

$x^2=ay$

$x^2=a\left(1+2x\right)$

$x^2-2ax-a=0$

$x=\frac{2a\pm \sqrt{4a^2+4a}}{2}$

$x=a\pm \sqrt{a^2+a}$

$y=1+2x$

$y=1+2\left(a\pm \sqrt{a^2+a}\right)$

$y=1+2a\pm 2\sqrt{a^2+a}$

$A\left(a+\sqrt{a^2+a},1+2a+2\sqrt{a^2+a}\right),B\left(a-\sqrt{a^2+a},1+2a-2\sqrt{a^2+a}\right)$

$AB=\sqrt{{\left(2\sqrt{a^2+a}\right)}^2+{\left(4\sqrt{a^2+a}\right)}^2}$

$\sqrt{40}=\sqrt{20\left(a^2+a\right)}$

$40=20\left(a^2+a\right)$

$a^2+a-2=0$

$a=1$