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  • The ellipse  <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D&plus;%5Cfrac%7By%5E2%7D%7B

The ellipse  \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}  and the circle\mathrm{ x^2+y^2=a}\mathrm{ x^2+y^2=ab} intersect at an angle

Option: 1

\mathrm{\theta=\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)}


Option: 2

\mathrm{\theta=\tan ^{-1}\left(\frac{a+b}{\sqrt{a b}}\right)}


Option: 3

\mathrm{\theta=\tan ^{-1}\left(\frac{1}{\sqrt{a b}}\right)}


Option: 4

\mathrm{\theta=\tan ^{-1}\left(\frac{a-b}{a b}\right)}


Answers (1)

best_answer

\mathrm{b^2 x^2+a^2 y^2=a^2 b^2 \quad \text { Ellipse }}    .... [1]

\mathrm{\text { and } x^2+y^2=a b \quad \text { Circle }}

\mathrm{\text { or } a^2 x^2+a^2 y^2=a^3 b}       ...[2]

Solving (1) and (2), by subtracting

\mathrm{\begin{aligned} & \left(b^2-a^2\right) x^2=a^2 b(b-a) \\ \therefore \quad & x^2=\frac{a^2 b}{b+a} \end{aligned}}

\mathrm{\text { Similarly, } y^2=\frac{a b^2}{a+b}}       ....[3]

\mathrm{\text { Above gives the point of intersection say }\left(x_1, y_1\right) \text {. }}

\mathrm{\text { Tanget at }\left(x_1, y_1\right) \text { to circle } x^2+y^2=a b \text { is }}

\mathrm{x x_1+y y_1=a b, m_1=-\frac{x_1}{y_1}=-\sqrt{\frac{a}{b}}}      (By (3))

Slope of tangent to ellipse is

\mathrm{-\frac{x_1}{y_1} \frac{b^2}{a^2}=-\sqrt{\frac{a}{b}} \cdot \frac{b^2}{a^2}=m_2 \text {, }}         (By (3))

\mathrm{\begin{aligned} & \tan \theta=\frac{m_1-m_2}{1+m_1 m_2}=\frac{\sqrt{\frac{a}{b}}\left[1-\frac{b^2}{a^2}\right]}{1+\frac{a}{b} \cdot \frac{b^2}{a^2}}=\frac{a^2-b^2}{a(a+b)} \cdot \sqrt{\frac{a}{b}} \\ & \text { or } \tan \theta=\frac{a-b}{\sqrt{a b}} . \end{aligned}}

 

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seema garhwal

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