The parabola x^2=4ay makes an intercept of le | vedclass

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(B) The equation of the parabola is $x^2=4ay$ and the line is $y=2x+1$. Here $m=2$ and $c=1$. Substituting $x = \frac{y-1}{2}$ into the parabola equat

Vedclass · Accepted Answer

$-2$

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(B) The equation of the parabola is $x^2=4ay$ and the line is $y=2x+1$. Here $m=2$ and $c=1$. Substituting $x = \frac{y-1}{2}$ into the parabola equation: $(\frac{y-1}{2})^2 = 4ay$ $\Rightarrow y^2 - 2y + 1 = 16ay$ $\Rightarrow y^2 - (16a+2)y + 1 = 0$. Let the roots be $y_1$ and $y_2$. Then $y_1+y_2 = 16a+2$ and $y_1y_2 = 1$. The points of intersection are $(x_1, y_1)$ and $(x_2, y_2)$. The length of the intercept is $L = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Since $y=2x+1$,$x_2-x_1 = \frac{y_2-y_1}{2}$. $L = \sqrt{\frac{5}{4}(y_2-y_1)^2} = \frac{\sqrt{5}}{2} \sqrt{(y_1+y_2)^2 - 4y_1y_2}$. Given $L = \sqrt{40}$,so $40 = \frac{5}{4} ((16a+2)^2 - 4)$. $32 = (16a+2)^2 - 4 \Rightarrow (16a+2)^2 = 36$. $16a+2 = 6$ or $16a+2 = -6$. $16a = 4$ $\Rightarrow a = 1/4$ $\Rightarrow 4a = 1$ (Not in options). $16a = -8$ $\Rightarrow a = -1/2$ $\Rightarrow 4a = -2$. Thus,the value of $4a$ is $-2$.

The parabola $x^2=4ay$ makes an intercept of length $\sqrt{40}$ units on the line $y=2x+1$. Find the value of $4a$.

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