If the line x+4y=√7, where R, touches the ellipse 3x^2+4y^2=1 at the point P in the first …

Question

If the line $\alpha x+4y=\sqrt{7}$, where $\alpha \in R$, touches the ellipse $3{x}^{2}+4{y}^{2}=1$ at the point P in the first quadrant, then one of the focal distances of $P$ is :

Accepted Answer

$\begin{gathered} \alpha x+4y-\sqrt{7}=03{x}^{2}+4{y}^{2}=1 \ 	herefore {c}^{2}={a}^{2}{m}^{2}+{b}^{2} \ \frac{7}{16}=\frac{1}{3}	imes \frac{{\alpha }^{2}}{16}+\frac{1}{4}\Rightarrow \alpha =3,-3 \end{gathered}$

Tangent is $3x+47-\sqrt{7}=0$

Let the point of contact is $P({x}_{1},{y}_{1})$

$\begin{gathered} 	herefore \frac{3{x}_{1}}{3}=\frac{4{y}_{1}}{4}=\frac{1}{\sqrt{7}} \ 	herefore P(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}}) \ =\sqrt{1-\frac{3}{4}}=\frac{1}{2} \end{gathered}$

$\begin{gathered} PS=e(PM)=e(\frac{a}{e}-\frac{1}{\sqrt{7}})=\frac{1}{2}(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{7}})=\frac{1}{\sqrt{3}}-\frac{1}{2\sqrt{7}} \ P{S}^{'}=e(P{M}^{'})=\frac{1}{2}(\frac{a}{e}+\frac{1}{\sqrt{7}})=\frac{1}{2}(\frac{1}{\sqrt{7}}+\frac{2}{\sqrt{3}})=\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}} \end{gathered}$

Solutions · 1