Let E be an ellipse whose axes are parallel to the co-ordinate axes, having its center at (3,-4) one focus at (4, – 4) and one vertex at (5,-4). If mx – y = 4, m > 0 is a tangent to the ellipse E, then the value of 5m² is equal to

Given $\mathrm{a} > \mathrm{b}$

$$\begin{gathered} \mathrm{centre}=(3, -4)=(\mathrm{h}, \mathrm{k}) and focus( S)=(4,-4) \\ \mathrm{CS}=\mathrm{ae}=1, \mathrm{CA}=2=\mathrm{a} \\ \Rightarrow \mathrm{e}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ Now {\mathrm{b}}^2={\mathrm{a}}^2(1-{\mathrm{e}}^2)=4\left(1-\frac{1}{4}\right) \\ =3 \\ \therefore \mathrm{eq}uation of ellipse is \frac{{(\mathrm{x}-3)}^2}{4}+\frac{{(\mathrm{y}+4)}^2}{3}=1 \end{gathered}$$

equation  of Tangent

$$\begin{gathered} \mathrm{y}+4=\mathrm{m}(\mathrm{x}-3)+\sqrt{4{\mathrm{m}}^2+3} \\ \mathrm{y}=\mathrm{mx}-3\mathrm{m}-4+\sqrt{4{\mathrm{m}}^2+3} — ① \end{gathered}$$

given of Tangent

$$\begin{gathered} \mathrm{y}=\mathrm{mx}-4 — ② \\ \mathrm{eq} ① \& ② \mathrm{same} \\ -4=-4-3\mathrm{m}+\sqrt{4{\mathrm{m}}^2+3} \\ 3\mathrm{m}=\sqrt{4{\mathrm{m}}^2+3} \\ \mathrm{S}.\mathrm{B}.\mathrm{S} \\ 9{\mathrm{m}}^2=4{\mathrm{m}}^2+3 \\ 5{\mathrm{m}}^2=3 \end{gathered}$$