$A B$ is a line segment moving between the axes such that ' $A$ ' lies on $X$-axis and ' $B$ ' lies on…

$A B$ 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
  1. $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
  2. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
  3. $\frac{x^2}{2 a^2}+\frac{y^2}{2 b^2}=1$
  4. $\frac{x^2}{2 b^2}+\frac{y^2}{a^2}=1$

Solution

$ \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

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

Practice more Ellipse questions on Aicharya