SATHEE: Ellipse 2 Question 19

Ellipse 2 Question 19

19. Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse of the point of contact meet on the corresponding directrix.

(2002, 5M)

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Answer:

Correct Answer: 19. (d)

Solution:

Any point on the ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 be P (a \cos θ, b \sin θ)$

The equation of tangent at point $P$ is given by

$\frac{x \cos θ}{a} + \frac{y \sin θ}{b} = 1$

The equation of line perpendicular to tangent is

$\frac{x \sin θ}{b} - \frac{y \cos θ}{a} = λ$

Since, it passes through the focus ( $a e, 0)$ , then

$\begin{array}{cc} \frac{a e \sin θ}{b} - 0 = λ \\ \Rightarrow λ = \frac{a e \sin θ}{b} \\ ∴ & Equation is \frac{x \sin θ}{b} - \frac{y \cos θ}{a} = \frac{a e \sin θ}{b} \end{array}$

Equation of line joining centre and point of contact $P (a \cos θ, b \sin θ)$ is

$y = \frac{b}{a} (\tan θ) x$

Point of intersection $Q$ of Eqs. (i) and (ii) has $x$ coordinate, $\frac{a}{e}$ . Hence, $Q$ lies on the corresponding directrix $x = \frac{a}{e}$ .

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Ellipse 2 Question 19

19. Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse of the point of contact meet on the corresponding directrix.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Ellipse 2 Question 19

19. Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse of the point of contact meet on the corresponding directrix.

Answer:

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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