If the length of the latus rectum of a parabo | vedclass

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(C) The distance from the focus $(a, a)$ to the tangent at the vertex $x+y-a=0$ is given by the formula for the distance from a point to a line:$d = \

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$4 \sqrt{2}$

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(C) The distance from the focus $(a, a)$ to the tangent at the vertex $x+y-a=0$ is given by the formula for the distance from a point to a line:$d = \frac{|a+a-a|}{\sqrt{1^2+1^2}} = \frac{|a|}{\sqrt{2}}$.This distance $d$ is equal to $a$ in the standard parabola equation $y^2 = 4ax$,where $4a$ is the length of the latus rectum.Given the length of the latus rectum is $16$,we have $4d = 16$,which implies $d = 4$.Therefore,$\frac{|a|}{\sqrt{2}} = 4$.$|a| = 4 \sqrt{2}$.

If the length of the latus rectum of a parabola,whose focus is $(a, a)$ and the tangent at its vertex is $x+y=a$,is $16$,then $|a|$ is equal to.

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