SATHEE: Ellipse 2 Question 9

Ellipse 2 Question 9

9. The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latusrectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ is

(2015 Main)

(a) $\frac{27}{4}$

(b) 18

(d) 27

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Solution:

Given equation of ellipse is

$\begin{array}{lc} \frac{x^{2}}{9} + \frac{y^{2}}{5} = 1 \\ ∴ & a^{2} = 9, b^{2} = 5 \Rightarrow a = 3, b = \sqrt{5} \\ Now, & e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{5}{9}} = \frac{2}{3} \end{array}$

Foci $= (\pm a e, 0) = (\pm 2, 0)$ and $\frac{b^{2}}{a} = \frac{5}{3}$

$∴$ Extremities of one of latusrectum are

$2, \frac{5}{3} and 2, \frac{- 5}{3}$

$∴$ Equation of tangent at $2, \frac{5}{3}$ is

$\frac{x (2)}{9} + \frac{y (5 / 3)}{5} = 1 or 2 x + 3 y = 9$

Since, Eq. (ii) intersects $X$ and $Y$ -axes at $\frac{9}{2}, 0$ and $(0, 3)$ , respectively.

$∴$ Area of quadrilateral $= 4 \times$ Area of $△ P O Q$

$= 4 \times \frac{1}{2} \times \frac{9}{2} \times 3 = 27 sq units$

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Ellipse 2 Question 9

9. The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latusrectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ is

Solution:

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सदस्यता

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Ellipse 2 Question 9

9. The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latusrectum to the ellipse x29+y25=1 is

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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9. The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latusrectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ is