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

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Aicharya · Accepted Answer

Since length of the chord intercepted from the line $y=m x+c$ by the parabola $x^2=4 a y$ is $4 \sqrt{a\left(1+m^2\right)\left(c+a m^2\right)}$ So, $\sqrt{40}=4 \sqrt{a(\mathrm{l}+4)(\mathrm{l}+a(4))}$ $\begin{array}{lc} \Rightarrow & 40=16(5 a(1+4 a)) \ \Rightarrow & 1=2 a(1+4 a) \ \Rightarrow & 8 a^2+2 a-1=0 \ \Rightarrow & 8 a^2+4 a-2 a-1=0 \ \Rightarrow & 4 a(2 a+1)-1(2 a+1)=0 \ \Rightarrow & a=\frac{1}{4} \text { or }-\frac{1}{2} \ \Rightarrow & 4 a=1 \text { or }-2 \end{array}$ Hence, option (2) is correct.

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

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