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Consider the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1\) (\(b>0\)). Let \(P_b=(1,\frac{\sqrt{3}b}{2})\) and let \(A\) be its projection on the x-axis, that is, \(A=(1,0)\).

(a) Verify that the point \(P_b\) lies on the ellipse and find the equation of the tangent line \(\ell_b\) at the ellipse at \(P_b.\)

(b) Note that as \(b\) varies, \(P_b\) varies, as well. Nevertheless, prove that the tangent(s) \(\ell_b\) drawn at the point(s) \(P_b\) cut the x-axis at the same point \(B\). [/centre]

(a) It is easy to verify that the point \(P_b\) lies on the ellipse. Now, by implicit differentiation we get
\[
\frac{2x}{4}+\frac{2y}{b^{2}}\frac{dy}{dx}=0
\]
and so
\[
\frac{dy}{dx}=-\frac{b^{2}}{4}\frac{x}{y}.
\]
Hence the tangent line at the ellipse at \(P_b=(1,\frac{\sqrt{3}b}{2})\) has equation
\[
\begin{align}
y &=-\frac{b}{2\sqrt{3}}(x-1)+\frac{\sqrt{3}b}{2}
=-\frac{b}{2\sqrt{3}}x+\frac{b}{2\sqrt{3}}+\frac{\sqrt{3}b}{2}\notag\\
&=-\frac{b}{2\sqrt{3}}x+\frac{2b}{\sqrt{3}} \notag.
\end{align}
\]
(b) For \(y=0\), we solve for \(x\) to get \(x=4\). Hence, the tangents \(\ell_b\) drawn at the point(s) \(P_b\) meet the \(x\)-axis at the same point \(B=(4,0)\).

Comment: For the general case, where the ellipse has equation
\[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\]
(\(b>0\) ) and \(P_b=(x_0,y_0)\), \( (x_0\ne0)\) we have \(B=(\frac{a^{2}}{x_0},0)\).