Let $E$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3,-4)

Question

Let $E$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3,-4)$, one focus at $(4,-4)$ and one vertex at $(5,-4)$. If $m x-y=4$, $m>0$ is a tangent to the ellipse $E$, then the value of $5 m^{2}$ is equal to

Professor PC Thomas Classes · Answer

(3) : The equation of ellipse is\ $\frac{(x-3)^{2}}{a^{2}}+\frac{(y+4)^{2}}{b^{2}}=1$ Now, vertex is $(5,-4)$\ $	herefore \quad \frac{(5-3)^{2}}{a^{2}}+0=1 \Rightarrow a^{2}=4$\ Now, $a-a e=5-4=1$\ $\Rightarrow a(1-e)=1 \Rightarrow 1-e=\frac{1}{2} \Rightarrow e=\frac{1}{2}$\ $\Rightarrow \quad \sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{1}{2} \Rightarrow 1-\frac{b^{2}}{4}=\frac{1}{4}$\ $\Rightarrow \quad 4-b^{2}=1 \Rightarrow b^{2}=3$\ $	herefore \quad$ Equation of ellipse is,\ $\frac{(x-3)^{2}}{4}+\frac{(y+4)^{2}}{3}=1$\ $	herefore \quad$ Now, equation of tangent is,\ $(y+4)=m(x-3) \pm \sqrt{4 m^{2}+3}$\ $\Rightarrow m x-y=4+3 m \pm \sqrt{4 m^{2}+3}$\ But given tangent is, $m x-y=4$\ $\Rightarrow 4=4+3 m \pm \sqrt{4 m^{2}+3}$\ $\Rightarrow-3 m= \pm \sqrt{4 m^{2}+3}$\ $\Rightarrow 9 m^{2}=4 m^{2}+3 \Rightarrow 5 m^{2}=3$\