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Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its centre 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 5 \mathrm{~m}^2 is equal to

Option: 1

4


Option: 2

5


Option: 3

3


Option: 4

2


Answers (1)

best_answer

:The equation of ellipse is

\mathrm{\frac{(x-3)^2}{a^2}+\frac{(y+4)^2}{b^2}=1}

Now, vertex is (5, –4)

\mathrm{\therefore \quad \frac{(5-3)^2}{a^2}+0=1 \Rightarrow a^2=4}

Distance between centre and focus is ae
ae = 1

\mathrm{\begin{aligned} & e=\frac{1}{a} \Rightarrow e=\frac{1}{2} \Rightarrow \sqrt{1-\frac{b^2}{a^2}}=\frac{1}{2} \Rightarrow 1-\frac{b^2}{4}=\frac{1}{4} \\ & \Rightarrow 4-b^2=1 \Rightarrow b^2=3 \end{aligned}}

\mathrm{\therefore \quad \text { Equation of ellipse is, } \frac{(x-3)^2}{4}+\frac{(y+4)^2}{3}=1}

Now, equation of tangent is,

\mathrm{(y+4)=m(x-3) \pm \sqrt{4 m^2+3}}

\mathrm{\Rightarrow \quad m x-y=4+3 m \pm \sqrt{4 m^2+3}}

But given tangent is, mx – y = 4

\mathrm{\begin{aligned} & \Rightarrow \quad 4=4+3 m \pm \sqrt{4 m^2+3} \\ & \Rightarrow \quad-3 m= \pm \sqrt{4 m^2+3} \\ & \Rightarrow \quad 9 m^2=4 m^2+3 \Rightarrow 5 m^2=3 \end{aligned}}

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Gaurav

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