A circle $x^2+y^2=4$ intersects the $x$-axis at $A(-2,0)$ and $B(2,0)$. If two variable points $P(2\cos\alpha,\,2\sin\alpha)$ and $Q(2\cos\beta,\,2\sin\beta)$ vary on the circle such that $\alpha-\beta=\dfrac{\pi}{2}$, then find the locus of the point of intersection of lines $AP$ and $BQ$.

Question

If eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which passes through the point $(3, 4)$ is $\frac{\sqrt{5}}{3}$, then length of latus rectum of ellipse, is :

Answer 1

To solve this problem, we need to find the locus of the point of intersection of lines $AP$ and $BQ$, given the conditions.

The circle given is x^2 + y^2 = 4, which implies it has a radius of 2 and is centered at the origin (0,0).

The circle intersects the $x$-axis at points $A(-2, 0)$ and $B(2, 0)$. Two variable points $P$ and $Q$ are given as $P(2\cos\alpha, 2\sin\alpha)$ and $Q(2\cos\beta, 2\sin\beta)$ respectively. Since $\alpha - \beta = \dfrac{\pi}{2}$, the points $P$ and $Q$ are orthogonal on the circle.

Let's parameterize the problem:

Point $P$ is defined as $P(2\cos\alpha, 2\sin\alpha)$.
Point $Q$ being orthogonally displaced by $\dfrac{\pi}{2}$ is defined as $Q(2\cos(\alpha - \frac{\pi}{2}), 2\sin(\alpha - \frac{\pi}{2}))$.

Using trigonometric identities:

$\cos(\alpha - \frac{\pi}{2}) = \sin\alpha$
$\sin(\alpha - \frac{\pi}{2}) = -\cos\alpha$

Thus, point $Q$ simplifies to $Q(2\sin\alpha, -2\cos\alpha)$.

Now, we find the equations of lines $AP$ and $BQ$:

For line $AP$, using the point-slope form, the slope = $\dfrac{2\sin\alpha - 0}{2\cos\alpha + 2} = \dfrac{\sin\alpha}{\cos\alpha + 1}$.
The equation of line $AP$ is therefore: $y - 0 = \dfrac{\sin\alpha}{\cos\alpha + 1}(x + 2)$.
For line $BQ$, the slope = $\dfrac{-2\cos\alpha - 0}{2\sin\alpha - 2} = \dfrac{-\cos\alpha}{\sin\alpha - 1}$.
The equation of line $BQ$ is: $y = \dfrac{-\cos\alpha}{\sin\alpha - 1}(x - 2)$.

To find the point of intersection of lines $AP$ and $BQ$, equate the equations:

$\dfrac{\sin\alpha}{\cos\alpha + 1}(x + 2) = \dfrac{-\cos\alpha}{\sin\alpha - 1}(x - 2)$.

Let the intersection point be $M(h, k)$. Solving these equations will give us the coordinates of $M$.

By simplifying, substituting back into circle's parametric coordinates and eliminating $\alpha$, we derive:

The locus is $x^2 + y^2 - 4y - 4 = 0$.

Thus, the correct answer is:

$x^2 + y^2 - 4y - 4 = 0$

Answer 2