The slope of a common tangent to the ellipse | vedclass

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(D) Let the equation of the common tangent be $y = mx + c$.For the ellipse $\frac{x^2}{49} + \frac{y^2}{4} = 1$,the condition for tangency is $c^2 = a

Vedclass · Accepted Answer

$\frac{2}{\sqrt{11}}$

Vedclass · Answer

(D) Let the equation of the common tangent be $y = mx + c$.For the ellipse $\frac{x^2}{49} + \frac{y^2}{4} = 1$,the condition for tangency is $c^2 = a^2m^2 + b^2$.Here $a^2 = 49$ and $b^2 = 4$,so $c^2 = 49m^2 + 4$ $(i)$.For the circle $x^2 + y^2 = 16$,the condition for tangency is $c^2 = r^2(1 + m^2)$.Here $r^2 = 16$,so $c^2 = 16(1 + m^2)$ (ii).Equating $(i)$ and (ii):$49m^2 + 4 = 16 + 16m^2$$33m^2 = 12$$m^2 = \frac{12}{33} = \frac{4}{11}$$m = \pm \frac{2}{\sqrt{11}}$Thus,the slope is $\frac{2}{\sqrt{11}}$.

The slope of a common tangent to the ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ and the circle $x^2+y^2=16$ is

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