The pole of the line 2x + 3y = 4 with respect | vedclass

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

Question

(A) Let the pole be $(x_1, y_1)$. The equation of the polar of the point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 = a^2$ is given by $xx_1 +

Vedclass · Accepted Answer

$(32, 48)$

Vedclass · Answer

(A) Let the pole be $(x_1, y_1)$. The equation of the polar of the point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 = a^2$ is given by $xx_1 + yy_1 = a^2$.Here,$a^2 = 64$,so the equation of the polar is $xx_1 + yy_1 = 64$ $... (i)$.Given that the line $2x + 3y = 4$ is the same as the polar line $xx_1 + yy_1 = 64$,we can write the equation as $xx_1 + yy_1 - 64 = 0$ and $2x + 3y - 4 = 0$.Comparing the coefficients of the two equations:$\frac{x_1}{2} = \frac{y_1}{3} = \frac{-64}{-4}$$\frac{x_1}{2} = 16 \Rightarrow x_1 = 32$$\frac{y_1}{3} = 16 \Rightarrow y_1 = 48$Thus,the pole is $(32, 48)$.

The pole of the line $2x + 3y = 4$ with respect to the circle $x^2 + y^2 = 64$ is:

Similar Questions

If $2kx + 3y - 1 = 0$ and $2x + y + 5 = 0$ are conjugate lines with respect to the circle $x^2 + y^2 - 2x - 4y - 4 = 0$,then $k =$

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 $(4,2)$ and $(k,-3)$ are conjugate points with respect to $x^2+y^2-5x+8y+6=0$,then $k$ equals

If the inverse point of the point $(-1, 1)$ with respect to the circle $x^2+y^2-2x+2y-1=0$ is $(p, q)$,then $p^2+q^2=$

The polars of $(-1, 2)$ with respect to the two circles $S_1 \equiv x^2+y^2+6y+7=0$ and $S_2 \equiv x^2+y^2+6x+1=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module