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)

$1$
$-2$

Any point on the parabola $x^2=ay$ can be written as $(x,\frac{x^2}{a})$.
Substituting in the equation of the line
$\frac{x^2}{a}-2x=1$
or
$x^2-2ax-a=0$
$x=\frac{2a\pm \sqrt{4a^2+4a}}{2}$
$x=a\pm \sqrt{a^2+a}$
Hence
$y=\frac{x^2}{a}=\frac{2a^2+a\pm 2a\sqrt{a^2+a}}{a}$
$=(2a+1)\pm 2\sqrt{a^2+a}$
Hence the points of intersection are
$A=(a+\sqrt{a^2+a},(2a+1)+2\sqrt{a^2+a})$ and $B=\left(\right(a-\sqrt{a^2+a},(2a+1)-2\sqrt{a^2+a})$
Applying distance formula
$A{B}^2=4(\sqrt{a^2+a}{)}^2+16(\sqrt{a^2+a}{)}^2=40$
$20(a^2+a)=40$
$a^2+a=2$
$a^2+a-2=0$
$(a+2)(a-1)=0$
$a=-2$ and $a=1$.