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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If the polar of a circle $x^2 + y^2 = a^2$ with respect to a point $(x', y')$ is $Ax + By + C = 0$,then its pole will be:

If $5x + 6y - 34 = 0$ and $2x + y + c = 0$ are conjugate lines with respect to the circle $x^2 + y^2 - 8x - 10y + 25 = 0$,then which of the following points lies on the line $2x + y + c = 0$?

The locus of the poles of the tangents to the circle $x^2+y^2-2x+2y-2=0$ with respect to the circle $x^2+y^2=4$ is:

For all real values of $k$,the point which lies on the polar of $(k, k+1)$ with respect to the circle $x^2+y^2+4x-8y-5=0$ is

The point of concurrence of all conjugate lines of the line $5x + 7y - 78 = 0$ with respect to the circle $x^2 + y^2 + 6x + 8y - 96 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module