$x - 2y + 4 = 0$ is a common tangent to $y^{2} = 4x$ and $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1$

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$x - 2y + 4 = 0$ is a common tangent to $y^{2} = 4x$ and $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1$. Then the value of $b$ and the other common tangent are given by :

Accepted Answer

Sol.(b) Equation of tangent at $P$ is$$\begin{array}{r} y = \frac{x}{2} + 2\#(1) \end{array}$$For (1) to be tangent to $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1$$${\Rightarrow \ (2)^{2} = 4\left( \frac{1}{2} \right)^{2} + b^{2} }{\Rightarrow \ b^{2} = 3\ \Rightarrow b = \pm \sqrt{3} }$$Equation of tangent at $Q$ is$$- 2y = x + 4 \Rightarrow 2y + x + 4 = 0$$

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