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The equation of the common tangent in \mathrm{1^{\text {st }}} quadrant to the circle \mathrm{x^2+y^2=16} and the ellipse \mathrm{\frac{x^2}{25}+\frac{y^2}{4}=1} is

Option: 1

\mathrm{y=-\frac{1}{\sqrt{3}} x+4 \sqrt{\frac{7}{3}}}


Option: 2

\mathrm{y=-\frac{2}{\sqrt{3}} x+\sqrt{\frac{7}{3}}}


Option: 3

\mathrm{y=\frac{1}{3} x+4 \sqrt{\frac{7}{3}}}


Option: 4

\mathrm{y=\frac{-2}{\sqrt{3}} x+4 \sqrt{\frac{7}{3}}}


Answers (1)

best_answer

Let the equations of tangents to the given circle and the ellipse respectively

\mathrm{y=m x+4 \sqrt{1+m^2}}                 .....[i]

\mathrm{\text { and } y=m x+\sqrt{25 m^2+4}}      ...[ii]

Since (i) and (ii) are coincident lines, so

\mathrm{\begin{aligned} & 4 \sqrt{1+m^2}=\sqrt{25 m^2+4} \\ & 16\left(1+m^2\right)=25 m^2+4 \end{aligned} \Rightarrow m= \pm \frac{2}{\sqrt{3}}}

m < 0 because common tangent in 1st quadrant
\mathrm{\text { So, } m=-\frac{2}{\sqrt{3}}}
So the equation of the common tangent is

\mathrm{y=-\frac{2}{\sqrt{3}} x+4 \sqrt{\frac{7}{3}}}              (From (i))

Posted by

Gautam harsolia

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