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Find the Equation of the Ellipse in the Case: (Iv) Focus is (1, 2), Directrix is 3x + 4y − 5 = 0 and E = 1 2 - Mathematics

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प्रश्न

Find the equation of the ellipse in the case: 

 focus is (1, 2), directrix is 3x + 4y − 5 = 0 and e = \[\frac{1}{2}\]

 

 

संक्षेप में उत्तर

उत्तर

\[\text{ Let S(1, 2) be the focus and ZZ' be the directrix . } \]
\[\text{ Let P(x, y) be any point on the ellipse and let PM be the perpendicular from P on the directrix .}  \]
\[\text{ Then by the definition, we have: } \]
\[SP = e \times PM\]
\[ \Rightarrow SP = \frac{1}{2} \times PM\]
\[ \Rightarrow 2SP = PM\]
\[ \Rightarrow 4 \left( SP \right)^2 = {PM}^2 \]
\[ \Rightarrow 4\left[ \left( x - 1 \right)^2 + \left( y - 2 \right)^2 \right] = \left| \frac{3x + 4y - 5}{\sqrt{3^2 + \left( 4 \right)^2}} \right|^2 \]
\[ \Rightarrow 4\left[ x^2 + 1 - 2x + y^2 + 4 - 4y \right] = \frac{9 x^2 + 16 y^2 + 25 + 24xy - 40y + 30x}{25}\]
\[ \Rightarrow 100 x^2 + 100 - 200x + 100 y^2 + 400 - 400y = 9 x^2 + 16 y^2 + 25 + 24xy - 40y - 30x\]
\[ \Rightarrow 91 x^2 + 84 y^2 - 24xy - 360y - 170x + 475 = 0\]
\[\text{ This is the required equation of the ellipse. } \]

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अध्याय 26: Ellipse - Exercise 26.1 [पृष्ठ २२]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 26 Ellipse
Exercise 26.1 | Q 2.4 | पृष्ठ २२

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