SATHEE: Ellipse 2 Question 3

Ellipse 2 Question 3

3. If the line $x - 2 y = 12$ is tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at the point $3, \frac{- 9}{2}$ , then the length of the latusrectum of the ellipse is

(2019 Main, 10 April I)

(a) $8 \sqrt{3}$

(b) 9

(d) $12 \sqrt{2}$

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Solution:

Key Idea Write equation of the tangent to the ellipse at any point and use formula for latusrectum of ellipse.

Equation of given ellipse is

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Now, equation of tangent at the point $3, - \frac{9}{2}$ on the ellipse (i) is

$\Rightarrow \frac{3 x}{a^{2}} - \frac{9 y}{2 b^{2}} = 1$

$[∵$ the equation of the tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at the point $(x_{1}, y_{1})$ is $\frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = 1]$

$∵$ Tangent (ii) represent the line $x - 2 y = 12$ , so

$\begin{aligned} \frac{1}{\frac{3}{a^{2}}} = \frac{2}{\frac{9}{2 b^{2}}} = \frac{12}{1} \\ \Rightarrow a^{2} = 36 and b^{2} = 27 \end{aligned}$

Now, Length of latusrectum $= \frac{2 b^{2}}{a} = \frac{2 \times 27}{6} = 9$ units

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Ellipse 2 Question 3

3. If the line $x - 2 y = 12$ is tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at the point $3, \frac{- 9}{2}$ , then the length of the latusrectum of the ellipse is

Solution:

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Ellipse 2 Question 3

3. If the line x−2y=12 is tangent to the ellipse x2a2+y2b2=1 at the point 3,−92, then the length of the latusrectum of the ellipse 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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3. If the line $x - 2 y = 12$ is tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at the point $3, \frac{- 9}{2}$ , then the length of the latusrectum of the ellipse is