If the parabola x^2=4ay, (a0) makes an inter | vedclass

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(A) The line $y=1+2x$ can be written as $2x-y+1=0$. Let the line intersect the parabola at points $A$ and $B$. The parametric form of the line passing

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(A) The line $y=1+2x$ can be written as $2x-y+1=0$. Let the line intersect the parabola at points $A$ and $B$. The parametric form of the line passing through $P(0,1)$ with slope $m=2$ is $\frac{x-0}{\cos 	heta} = \frac{y-1}{\sin 	heta} = r$,where $	an 	heta = 2$. Thus,$\cos 	heta = \frac{1}{\sqrt{5}}$ and $\sin 	heta = \frac{2}{\sqrt{5}}$.Substituting $x = \frac{r}{\sqrt{5}}$ and $y = 1 + \frac{2r}{\sqrt{5}}$ into the parabola equation $x^2=4ay$:$\left(\frac{r}{\sqrt{5}}ight)^2 = 4a\left(1 + \frac{2r}{\sqrt{5}}ight)$$\frac{r^2}{5} = 4a + \frac{8ar}{\sqrt{5}}$$r^2 - 8\sqrt{5}ar - 20a = 0$Let $r_1$ and $r_2$ be the roots of this quadratic equation. The length of the intercept is $|r_1 - r_2| = \sqrt{40}$.Using $(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1r_2$:$40 = (8\sqrt{5}a)^2 - 4(-20a)$$40 = 320a^2 + 80a$$32a^2 + 8a - 4 = 0$$8a^2 + 2a - 1 = 0$$(4a-1)(2a+1) = 0$Since $a>0$,we have $4a=1$.

If the parabola $x^2=4ay, (a>0)$ makes an intercept of length $\sqrt{40}$ units on the line $y=1+2x$,then $4a=$

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