Question
In the line $\alpha x + 4y = \sqrt{7}$, where $\alpha \in R$, touches the ellipse $3x^{2} + 4y^{2} = 1$ at the point $P$ in the first quadrant, then one of the focal distances of P is :
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Solution
$\alpha x + 4y - \sqrt{7} = 0$ touches $3x^{2} + 4y^{2} = 1$
$${\therefore c^{2} = a^{2}{\text{ }m}^{2} + b^{2} }{\frac{7}{16} = \frac{1}{3} \times \frac{\alpha^{2}}{16} + \frac{1}{4} \Rightarrow \alpha = 3, - 3 }$$Tangent is $3x + 4y - \sqrt{7} = 0$
Let the point of contact is $P\left( x_{1}y_{1} \right)$
∴ Tangent is $3{xx}_{1} + 4{yy}_{1} = 1$
$${\therefore\frac{3x_{1}}{3} = \frac{4y_{1}}{4} = \frac{1}{\sqrt{7}}\ \therefore P\left( \frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}} \right) }{e = \sqrt{1 - \frac{3}{4}} = \frac{1}{2} }$$
$${PS = e(PM) }{= e\left( \frac{a}{e} - \frac{1}{\sqrt{7}} \right) }{= \frac{1}{2}\left( \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{7}} \right) = \frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{7}} }{{PS}' = e\left( {PM}' \right) = \frac{1}{2}\left( \frac{a}{e} + \frac{1}{\sqrt{7}} \right) = \frac{1}{2}\left( \frac{1}{\sqrt{7}} + \frac{2}{\sqrt{3}} \right) }{= \frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}}$$
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