The shortest distance from (0,3) to the parab | vedclass

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(B) Let a point $B$ on the parabola be $B\left(\frac{K^2}{4}, K\right)$ and $A$ be $(0,3)$.The distance $AB = \sqrt{\left(\frac{K^2}{4} - 0\right)^2 +

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(B) Let a point $B$ on the parabola be $B\left(\frac{K^2}{4}, K\right)$ and $A$ be $(0,3)$.The distance $AB = \sqrt{\left(\frac{K^2}{4} - 0\right)^2 + (K - 3)^2} = \sqrt{\frac{K^4}{16} + K^2 - 6K + 9}$.Let $f(K) = AB^2 = \frac{K^4}{16} + K^2 - 6K + 9$.To find the shortest distance,we minimize $f(K)$ by setting $f'(K) = 0$.$f'(K) = \frac{4K^3}{16} + 2K - 6 = \frac{K^3}{4} + 2K - 6 = 0$.Multiplying by $4$,we get $K^3 + 8K - 24 = 0$.By inspection,$K=2$ is a root: $(2)^3 + 8(2) - 24 = 8 + 16 - 24 = 0$.Dividing $K^3 + 8K - 24$ by $(K-2)$,we get $(K-2)(K^2 + 2K + 12) = 0$.The quadratic $K^2 + 2K + 12$ has a negative discriminant $(D = 4 - 48 = -44)$,so $K=2$ is the only real solution.At $K=2$,the point $B$ is $\left(\frac{2^2}{4}, 2\right) = (1, 2)$.The shortest distance $AB = \sqrt{(1-0)^2 + (2-3)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.

List-$I$	List-$II$
$I$. Vertex	$A$. $8$
$II$. Length of latus rectum	$B$. $(\frac{29}{10}, \frac{-38}{10})$
$III$. Directrix	$C$. $3x+4y-1=0$
$IV$. One end of the latus rectum	$D$. $(\frac{-2}{5}, \frac{-16}{5})$
	$E$. $6$

The shortest distance from $(0,3)$ to the parabola $y^2=4x$ is

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