The question is: The hyperbola $H$ has equation $\frac{x^2}{16} - \frac{y^2}{4}=1$. The line $l_1$ is the tangent to $H$ at the point $P(4\sec(t), 2\tan(t))$
(a) Use calculus to show that an equation of $l_1$ is $2y\sin(t)=x-4\cos(t)$
What I have done for this part:
$x=4\sec(t) \Longrightarrow \frac{dx}{dt} = 4\sec(t)\tan(t) = \frac{4\sin(t)}{\cos^2(t)}$
$y=2\tan(t) \Longrightarrow \frac{dy}{dt} = 2\sec^2(t)=\frac{2}{\cos^2(t)}$
$\therefore \frac{dy}{dx}=\frac{1}{2\sin(t)}$
Let equation of tangent be $y-y_1=m(x-x_1)$
$$ \Leftrightarrow y-2\tan(t)=\frac{1}{2\sin(t)}(x-4\sec(t))$$
$$ \Leftrightarrow y-\frac{2\sin(t)}{\cos(t)} = \frac{1}{2\sin(t)}(x-\frac{4}{\cos(t)})$$
$$ \Leftrightarrow y-\frac{2\sin(t)}{\cos(t)} =\frac{x}{2\sin(t)}-\frac{2}{\sin(t)\cos(t)}$$
$$2y\sin(t)\cos(t)-4\sin^2(t)=x\cos(t)-4$$
$$2y\sin(t)\cos(t)=x\cos(t)+4(\sin^2(t)-1)$$
$$2y\sin(t)\cos(t)=x\cos(t)-4\cos^2(t) $$
$$2y\sin(t)=x-4\cos(t) $$
As desired
The part where I am stuck is this part:
(b) The line $l_2$ passes through origin and is perpendicular to $l_1$.
The lines $l_1$ and $l_2$ intersect at the point $Q$.
Show that, as $t$ varies, an equation of the locus of $Q$ is
$$(x^2+y^2)^2 = 16x^2-4y^2 $$
What I have tried for this part
Since line 2 passes through the origin its equation is just $y=mx$
Since it is perpendicular to $l_1$ I think it means $m_1 \cdot m_2 = -1$
So $ m_2 = \frac{-1}{\frac{1}{2\sin(t)}} = -2\sin(t)$
So would the equation of this line be $y=-2\sin(t)$ ?
Now I am confused how to continue
The equation for the second line is $y=-2\sin(t) x$. The point $Q$ satisfies equations for both lines. So you take $\sin(t)=-\frac{y}{2x}$ and plug it into the left side of equation 1. You get $$-\frac{y^2}{x}=x-4\cos(t)$$ You can multiply with $-x$, then move the terms without $t$ on the left hand side. $$x^2+y^2=4x\cos(t)$$ You now square this equation, use $\cos^2(t)=1-\sin^2(t)=1-\frac{y^2}{4x^2}$ and you should get the result in the problem