A line of fixed length a + b, where a neq b, | vedclass

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(C) Let the two perpendicular lines be the coordinate axes. Let the ends of the line segment be $A(0, m)$ and $B(n, 0)$.The length of the line segment

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an ellipse

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(C) Let the two perpendicular lines be the coordinate axes. Let the ends of the line segment be $A(0, m)$ and $B(n, 0)$.The length of the line segment is $AB = \sqrt{m^2 + n^2} = a + b$, so $m^2 + n^2 = (a + b)^2$.Let $P(h, k)$ be a point on the line segment $AB$ that divides it into parts of lengths $a$ and $b$. By the section formula, $P$ divides $AB$ in the ratio $b : a$ (since $AP = a$ and $PB = b$).Thus, $h = \frac{b(0) + a(n)}{a + b} = \frac{an}{a + b} \Rightarrow n = \frac{(a + b)h}{a}$.And $k = \frac{b(m) + a(0)}{a + b} = \frac{bm}{a + b} \Rightarrow m = \frac{(a + b)k}{b}$.Substituting these into $m^2 + n^2 = (a + b)^2$, we get:$\left(\frac{(a + b)k}{b}\right)^2 + \left(\frac{(a + b)h}{a}\right)^2 = (a + b)^2$.Dividing by $(a + b)^2$, we obtain $\frac{k^2}{b^2} + \frac{h^2}{a^2} = 1$.Replacing $(h, k)$ with $(x, y)$, the locus is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which is an ellipse.

$A$ line of fixed length $a + b$, where $a \neq b$, moves such that its ends are always on two fixed perpendicular straight lines. The locus of a point which divides the line into two parts of lengths $a$ and $b$ is

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