A line touches the ellipse x2-a2-y2-b2-1 and the circle x2-y2-r2 Then the slope m of the common tangent is given by m2-

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Eccentric Angle : Ellipse

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Question

$A$ line touches the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and the circle $x^{2} + y^{2} = r^{2}$ Then the slope $m$ of the common tangent is given by $m^{2} =$

\frac{a^{2} - r^{2}}{b^{2} - r^{2}}

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\frac{r^{2} - b^{2}}{a^{2} - r^{2}}

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\frac{r^{2} + b^{2}}{a^{2} - r^{2}}

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\frac{r^{2} - 2 b^{2}}{a^{2} - 2 r^{2}}

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Solution

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r (< a)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)

(a,b) is the mid point of the chord $¯ A B$ of the circle $x^{2} + y^{2} = r^{2}$ . The tangent at A,B meet a C. then area of $Δ A B C$

(x - a)^{2} + (y - b)^{2} = r^{2}, (x - b)^{2} + (y - a)^{2} = r^{2}

y = m x + c

(x - a)^{2} + (y - b)^{2} = r^{2}

m^{2} (a^{2} - r^{2}) + 2 m a (c - b) + (c - b)^{2} = r^{2}

The length of the chord intercepted by the circle $x^{2} + y^{2} = r^{2}$ on the line $\frac{x}{a} + \frac{y}{b} = 1$

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