Question

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

• A. $1$
• B. $-2$
• C. $-1$
• D. $2$

Answer 1

Answer: (A), (B)

$-2$

Solving the equations $x^2=ay$ and $y-2x=1$, we get
$x^2=a(2x+1)$
$\Rightarrow x^2-2ax-a=0$
So, the given line cuts the parabola at two points $(x_1,y_1)$ and $(x_2,y_2)$
$\therefore x_1+x_2=2a\text{and}{x}_1{x}_2=-a$

Now, $(\sqrt{40}{)}^2=(x_1-x_2{)}^2+(y_1-y_2{)}^2\Rightarrow 40=(x_1-x_2{)}^2+{(\frac{x_1^2}{a}-\frac{x_2^2}{a})}^2\Rightarrow 40=(x_1-x_2{)}^2[1+\frac{(x_1+x_2{)}^2}{a^2}]\Rightarrow 40=[(x_1+x_2{)}^2-4x_1{x}_2](\frac{4a^2}{a^2}+1)\Rightarrow 40=5(4a^2+4a)\Rightarrow a^2+a-2=0$
$\Rightarrow (a+2)(a-1)=0\therefore a=1,-2$