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Ellipse

Section: Mathematics

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Question : 6 of 16

Marks: +1, -0

The square of the slope of a common tangent drawn to the circle

4 x^{2}+4 y^{2}=25

and the ellipse

4 x^{2}+9 y^{2}=36

[AP EAPCET 21 May 2025 S2]

1
$\;\frac{2}{3}$
$\;{2}/{3}$
2

Solution:

We have a circle

x^{2}+y^{2}=\;\frac{25}{4}

and Ellipse

\;\frac{x^{2}}{9}+\;\frac{y^{2}}{4}=1

Let equation of tangent to the ellipse be

y=m x \pm \sqrt{9 m^{2}+4}

Given this tangent also touches the circle

\because

Condition of tendency

\;\frac{\sqrt{9 m^{2}+4}}{\sqrt{1+m^{2}}}=\;\frac{5}{2}

On squaring both sides, we get

\;4(9 m^{2}+4)=25(1+m^{2})

\Rightarrow 36 m^{2}+16=25+25 m^{2}

\Rightarrow 11 m^{2}=9

\therefore \;m^{2}=\;\frac{9}{11}

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