A(a, 0) and B(-a, 0) are two fixed points. If | vedclass

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

Question

(D) Let $C = (h, k)$.In $	riangle CDA$,$	an A = \frac{k}{a-h}$.In $	riangle CDB$,$	an B = \frac{k}{h-(-a)} = \frac{k}{h+a}$.Given $\angle A - \ang

Vedclass · Accepted Answer

$x^2 - y^2 + 2xy \cot 	heta = a^2$

Vedclass · Answer

(D) Let $C = (h, k)$.In $	riangle CDA$,$	an A = \frac{k}{a-h}$.In $	riangle CDB$,$	an B = \frac{k}{h-(-a)} = \frac{k}{h+a}$.Given $\angle A - \angle B = 	heta$,so $	an(A - B) = 	an 	heta$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B} = 	an 	heta$,$\frac{\frac{k}{a-h} - \frac{k}{h+a}}{1 + \frac{k}{a-h} \cdot \frac{k}{h+a}} = 	an 	heta$.$\frac{k(h+a) - k(a-h)}{(a-h)(h+a) + k^2} = 	an 	heta$.$\frac{2kh}{a^2 - h^2 + k^2} = 	an 	heta$.$2kh = (a^2 - h^2 + k^2) 	an 	heta$.$2kh \cot 	heta = a^2 - h^2 + k^2$.$h^2 - k^2 + 2hk \cot 	heta = a^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - y^2 + 2xy \cot 	heta = a^2$.

$A(a, 0)$ and $B(-a, 0)$ are two fixed points. If $\angle A - \angle B = \theta$,what is the locus of point $C$ of triangle $ABC$?

Similar Questions

If a point $(x, y) \equiv (\tan \theta + \sin \theta, \tan \theta - \sin \theta)$,then the locus of $(x, y)$ is

If the distance of any point $P(x, y)$ from the points $A(a + b, a - b)$ and $B(a - b, a + b)$ are equal,then the locus of $P$ is:

If a straight line drawn through the point of intersection of the lines $4x + 3y - 1 = 0$ and $3x + 4y - 1 = 0$ meets the coordinate axes at the points $P$ and $Q$, then the locus of the midpoint of $PQ$ is:

If the algebraic sum of the perpendicular distances from the points $(2,0)$,$(0,2)$,and $(1,1)$ to a variable line is zero,then the variable line always passes through a fixed point. The coordinates of that point are

If the sum of distances from a point $P$ to two mutually perpendicular straight lines is $1$ unit, then the locus of $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module