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$A B$ 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 $A B$ such that $P A=b$ and $P B=a$, then the equation of locus of $P$ is
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Verified Answer
The correct answer is:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$$
\text { Let } P(h, k) \text { be any point in the locus } \mathrm{cm}
$$
Let $A=(x, 0), B=(0, y)$
Given, $P A=b, P B=a$
In $\triangle P M A$,
$$
\sin \theta=\frac{k}{b}
$$
In $\triangle B N P$,
$$
\cos \theta=\frac{h}{a}
$$
We have,
$$
\begin{aligned}
\sin ^2 \theta+\cos ^2 \theta & =1 \\
\frac{k^2}{b^2}+\frac{h^2}{a^2} & =1 \\
\therefore \quad \frac{h^2}{a^2}+\frac{k^2}{b^2} & =1
\end{aligned}
$$
Required locus of point $P(h, k)$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Hence, option (2) is correct.
\text { Let } P(h, k) \text { be any point in the locus } \mathrm{cm}
$$
Let $A=(x, 0), B=(0, y)$
Given, $P A=b, P B=a$
In $\triangle P M A$,
$$
\sin \theta=\frac{k}{b}
$$
In $\triangle B N P$,
$$
\cos \theta=\frac{h}{a}
$$
We have,
$$
\begin{aligned}
\sin ^2 \theta+\cos ^2 \theta & =1 \\
\frac{k^2}{b^2}+\frac{h^2}{a^2} & =1 \\
\therefore \quad \frac{h^2}{a^2}+\frac{k^2}{b^2} & =1
\end{aligned}
$$
Required locus of point $P(h, k)$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Hence, option (2) is correct.
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