Question:
The length of the latus rectum of the parabola
\[
(x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2
\]
is
The length of the latus rectum of the parabola
\[
(x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2
\]
is
Show Hint
If a parabola is given using focus-directrix form, then first identify the focus and directrix. The distance between focus and directrix is \(2a\), and the length of the latus rectum is \(4a\).
Updated On: Jun 26, 2026
- \(\frac{1}{5}\)
- \(\frac{2}{5}\)
- \(\frac{3}{5}\)
- \(\frac{4}{5}\)
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The Correct Option is B
Solution and Explanation
Step 1: Compare the given equation with parabola definition.
The given equation is \[ (x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2 \] This can be written as \[ \sqrt{(x-2)^2+(y-3)^2} = \frac{|3x-4y+7|}{5} \] Now, \[ \frac{|3x-4y+7|}{\sqrt{3^2+(-4)^2}} = \frac{|3x-4y+7|}{5} \] So, the equation represents the set of points whose distance from the point \[ (2,3) \] is equal to their distance from the line \[ 3x-4y+7=0 \] Hence, the focus is \[ S(2,3) \] and the directrix is \[ 3x-4y+7=0 \]
Step 2: Find the distance between focus and directrix.
The distance of point \((2,3)\) from the line \(3x-4y+7=0\) is \[ d=\frac{|3(2)-4(3)+7|}{\sqrt{3^2+(-4)^2}} \] \[ d=\frac{|6-12+7|}{5} \] \[ d=\frac{|1|}{5} \] \[ d=\frac{1}{5} \]
Step 3: Use the relation between focus-directrix distance and latus rectum.
For a parabola, the distance between the focus and the directrix is \[ 2a \] Here, \[ 2a=\frac{1}{5} \] So, \[ a=\frac{1}{10} \] The length of the latus rectum of a parabola is \[ 4a \] Therefore, \[ 4a=4\left(\frac{1}{10}\right) \] \[ 4a=\frac{2}{5} \]
Step 4: Final conclusion.
Hence, the length of the latus rectum is \[ \boxed{\frac{2}{5}} \]
The given equation is \[ (x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2 \] This can be written as \[ \sqrt{(x-2)^2+(y-3)^2} = \frac{|3x-4y+7|}{5} \] Now, \[ \frac{|3x-4y+7|}{\sqrt{3^2+(-4)^2}} = \frac{|3x-4y+7|}{5} \] So, the equation represents the set of points whose distance from the point \[ (2,3) \] is equal to their distance from the line \[ 3x-4y+7=0 \] Hence, the focus is \[ S(2,3) \] and the directrix is \[ 3x-4y+7=0 \]
Step 2: Find the distance between focus and directrix.
The distance of point \((2,3)\) from the line \(3x-4y+7=0\) is \[ d=\frac{|3(2)-4(3)+7|}{\sqrt{3^2+(-4)^2}} \] \[ d=\frac{|6-12+7|}{5} \] \[ d=\frac{|1|}{5} \] \[ d=\frac{1}{5} \]
Step 3: Use the relation between focus-directrix distance and latus rectum.
For a parabola, the distance between the focus and the directrix is \[ 2a \] Here, \[ 2a=\frac{1}{5} \] So, \[ a=\frac{1}{10} \] The length of the latus rectum of a parabola is \[ 4a \] Therefore, \[ 4a=4\left(\frac{1}{10}\right) \] \[ 4a=\frac{2}{5} \]
Step 4: Final conclusion.
Hence, the length of the latus rectum is \[ \boxed{\frac{2}{5}} \]
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