The locus of the poles of normal chords of an | vedclass

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

Question

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(i)$.Let $(h, k)$ be the pole of the chord.The equation of the polar o

Vedclass · Accepted Answer

$\frac{a^6}{x^2} + \frac{b^6}{y^2} = (a^2 - b^2)^2$

Vedclass · Answer

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(i)$.Let $(h, k)$ be the pole of the chord.The equation of the polar of $(h, k)$ with respect to the ellipse is $\frac{xh}{a^2} + \frac{yk}{b^2} = 1$ $(ii)$.If this chord is a normal at point $	heta$,its equation is $ax \sec 	heta - by \csc 	heta = a^2 - b^2$ $(iii)$.Comparing $(ii)$ and $(iii)$,we have:$\frac{h/a^2}{a \sec 	heta} = \frac{k/b^2}{-b \csc 	heta} = \frac{1}{a^2 - b^2}$.This gives $\cos 	heta = \frac{a^3}{h(a^2 - b^2)}$ and $\sin 	heta = \frac{-b^3}{k(a^2 - b^2)}$.Using $\cos^2 	heta + \sin^2 	heta = 1$,we get:$\frac{a^6}{h^2(a^2 - b^2)^2} + \frac{b^6}{k^2(a^2 - b^2)^2} = 1$.Thus,the locus of $(h, k)$ is $\frac{a^6}{x^2} + \frac{b^6}{y^2} = (a^2 - b^2)^2$.

The locus of the poles of normal chords of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by:

Similar Questions

The lines $x \cos \alpha + y \sin \alpha = P, \alpha \in R$ are chords of the hyperbola $\frac{x^2}{9} - \frac{y^2}{36} = 1$ and they subtend a right angle at the centre of the hyperbola. The locus of the poles of these lines with respect to the given hyperbola is

If the pole of the line $x+2by-5=0$ with respect to the circle $S \equiv x^2+y^2-4x-6y+4=0$ lies on the line $x+by+1=0$,then the polar of the point $(b,-b)$ with respect to the circle $S=0$ is

The coordinates of the pole of the line $lx + my + n = 0$ with respect to the circle $x^2 + y^2 = 1$ are:

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

If the poles of the line $x-y=0$ with respect to the circles $x^2+y^2-2g_ix+c_i^2=0$ $(i=1, 2, 3)$ are $(\alpha_i, \beta_i)$,then $\sum_{i=1}^3 \frac{\alpha_i+\beta_i}{g_i}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module