AB is a line segment moving between the axes | vedclass

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(B) Let $P(h, k)$ be any point on the locus. Let $A$ be on the $X$-axis and $B$ be on the $Y$-axis. Let $	heta$ be the angle that the line segment $A

Vedclass · Accepted Answer

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Vedclass · Answer

(B) Let $P(h, k)$ be any point on the locus. Let $A$ be on the $X$-axis and $B$ be on the $Y$-axis. Let $	heta$ be the angle that the line segment $AB$ makes with the $X$-axis. From the geometry of the figure,in $	riangle PMA$,we have $\sin 	heta = \frac{k}{b}$,which implies $k = b \sin 	heta$. In $	riangle BNP$,we have $\cos 	heta = \frac{h}{a}$,which implies $h = a \cos 	heta$. Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we substitute the values: $(\frac{k}{b})^2 + (\frac{h}{a})^2 = 1$. Replacing $(h, k)$ with $(x, y)$,the equation of the locus is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

$AB$ is a line segment moving between the axes such that '$A$' lies on $X$-axis and '$B$' lies on $Y$-axis. If $P$ is a point on $AB$ such that $PA=b$ and $PB=a$,then the equation of the locus of $P$ is

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