The line x-2y-3=0 cuts the parabola y^2=4ax a | vedclass

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(A) The standard form of the parabola is $y^2=4ax$. The focus is given as $(a, 0) = (\frac{1}{4}, k)$.Comparing these,we get $a = \frac{1}{4}$ and $k

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$16a\sqrt{5}$

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(A) The standard form of the parabola is $y^2=4ax$. The focus is given as $(a, 0) = (\frac{1}{4}, k)$.Comparing these,we get $a = \frac{1}{4}$ and $k = 0$.So,the parabola is $y^2 = 4(\frac{1}{4})x$,which simplifies to $y^2 = x$.The line is $x - 2y - 3 = 0$,so $x = 2y + 3$.Substituting $x$ in the parabola equation: $y^2 = 2y + 3 \Rightarrow y^2 - 2y - 3 = 0$.Factoring the quadratic: $(y-3)(y+1) = 0$,so $y = 3$ or $y = -1$.If $y = 3$,$x = 2(3) + 3 = 9$. So $Q = (9, 3)$.If $y = -1$,$x = 2(-1) + 3 = 1$. So $P = (1, -1)$.The distance $PQ = \sqrt{(9-1)^2 + (3-(-1))^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$.Since $a = \frac{1}{4}$,we have $16a = 16(\frac{1}{4}) = 4$.Therefore,$PQ = 16a\sqrt{5}$.

The line $x-2y-3=0$ cuts the parabola $y^2=4ax$ at the points $P$ and $Q$. If the focus of this parabola is $(\frac{1}{4}, k)$,then $PQ=$

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