If the parabola x^{2}=ay makes an intercept o | vedclass

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(A, B) Given, the equation of the parabola is $x^{2}=ay$, which implies $y=\frac{x^{2}}{a}$.The equation of the line is $y=2x+1$.Substituting $y$ in t

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(A, B) Given, the equation of the parabola is $x^{2}=ay$, which implies $y=\frac{x^{2}}{a}$.The equation of the line is $y=2x+1$.Substituting $y$ in the parabola equation: $\frac{x^{2}}{a}=2x+1 \Rightarrow x^{2}-2ax-a=0$.Let the roots be $x_{1}$ and $x_{2}$. Then $x_{1}+x_{2}=2a$ and $x_{1}x_{2}=-a$.The difference $|x_{1}-x_{2}| = \sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = \sqrt{4a^{2}+4a} = 2\sqrt{a^{2}+a}$.Since the points lie on $y=2x+1$, the distance $d$ between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.Since $y_{2}-y_{1} = 2(x_{2}-x_{1})$, we have $d = \sqrt{(x_{2}-x_{1})^{2}+4(x_{2}-x_{1})^{2}} = |x_{2}-x_{1}|\sqrt{5}$.Given $d=\sqrt{40}$, so $\sqrt{40} = 2\sqrt{a^{2}+a} \cdot \sqrt{5} \Rightarrow \sqrt{40} = \sqrt{20(a^{2}+a)}$.Squaring both sides: $40 = 20(a^{2}+a) \Rightarrow a^{2}+a-2=0$.$(a+2)(a-1)=0$, so $a=1$ or $a=-2$.

List-$I$	List-$II$
$(I)$ Vertex	$(A)$ $(-\frac{3}{2}, -3)$
$(II)$ Focus	$(B)$ $(\frac{3}{2}, -3)$
$(III)$ Equation of the directrix	$(C)$ $2x+5=0$
$(IV)$ Equation of the axis	$(D)$ $2x+y+3=0$
	$(E)$ $y+3=0$
	$(F)$ $(-2, -3)$

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

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