The pole of the line frac{x}{a} + frac{y}{b} | vedclass

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(A) Let $P(x_1, y_1)$ be the pole with respect to the circle $x^2 + y^2 = c^2$. The equation of the polar of $P$ is given by $T = 0$,which is $x x_1 +

Vedclass · Accepted Answer

$\left(\frac{c^2}{a}, \frac{c^2}{b}\right)$

Vedclass · Answer

(A) Let $P(x_1, y_1)$ be the pole with respect to the circle $x^2 + y^2 = c^2$. The equation of the polar of $P$ is given by $T = 0$,which is $x x_1 + y y_1 = c^2$. This can be rewritten as $\frac{x x_1}{c^2} + \frac{y y_1}{c^2} = 1$. Comparing this with the given line equation $\frac{x}{a} + \frac{y}{b} = 1$,we get: $\frac{x_1}{c^2} = \frac{1}{a} \Rightarrow x_1 = \frac{c^2}{a}$ $\frac{y_1}{c^2} = \frac{1}{b} \Rightarrow y_1 = \frac{c^2}{b}$ Thus,the pole is $\left(\frac{c^2}{a}, \frac{c^2}{b}\right)$.

The pole of the line $\frac{x}{a} + \frac{y}{b} = 1$ with respect to the circle $x^2 + y^2 = c^2$ is

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