SATHEE: Ellipse 2 Question 1

Ellipse 2 Question 1

1. If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point $P$ on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at $P$ passes through $Q (4, 4)$ then $P Q$ is equal to

(a) $\frac{5 \sqrt{5}}{2}$

(b) $\frac{\sqrt{61}}{2}$

(d) $\frac{\sqrt{157}}{2}$

(2019 Main, 12 April I)

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Solution:

Key Idea Equation of tangent and normal to the ellipse

$\begin{aligned} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at point p (x_{1}, y_{1}) is T = 0 \Rightarrow \frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = 1 \\ and \frac{a^{2} x}{x_{1}} - \frac{b^{2} y}{y_{1}} = a^{2} - b^{2} respectively. \end{aligned}$

Equation of given ellipse is $3 x^{2} + 4 y^{2} = 12$

$\Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$

Now, let point $P (2 \cos θ, \sqrt{3} \sin θ)$ , so equation of tangent to ellipse (i) at point $P$ is

$\frac{x \cos θ}{2} + \frac{y \sin θ}{\sqrt{3}} = 1$

Since, tangent (ii) passes through point $Q (4, 4)$

$∴ 2 \cos θ + \frac{4}{\sqrt{3}} \sin θ = 1$

and equation of normal to ellipse (i) at point $P$ is

$\begin{aligned} \frac{4 x}{2 \cos θ} - \frac{3 y}{\sqrt{3} \sin θ} = 4 - 3 \\ \Rightarrow & 2 x \sin θ - \sqrt{3} \cos θ y = \sin θ \cos θ \end{aligned}$

Since, normal (iv) is parallel to line, $2 x + y = 4$

$∴$ Slope of normal (iv) $=$ slope of line, $2 x + y = 4$

$\Rightarrow \frac{2}{\sqrt{3}} \tan θ = - 2 \Rightarrow \tan θ = - \sqrt{3} \Rightarrow θ = 120^{\circ}$

$\Rightarrow (\sin θ, \cos θ) = \frac{\sqrt{3}}{2}, - \frac{1}{2}$

Hence, point $P - 1, \frac{3}{2}$

Now, $P Q = \sqrt{(4 + 1)^{2} + 4 - {\frac{3}{2}}^{2}}$

$= \sqrt{25 + \frac{25}{4}} = \frac{5 \sqrt{5}}{2}$

[given cordinates of $Q \equiv (4, 4)$ ]

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Ellipse 2 Question 1

1. If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point $P$ on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at $P$ passes through $Q (4, 4)$ then $P Q$ is equal to

Solution:

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एनसीईआरटी व्यायाम वीडियो समाधान

Ellipse 2 Question 1

1. If the normal to the ellipse 3x2+4y2=12 at a point P on it is parallel to the line, 2x+y=4 and the tangent to the ellipse at P passes through Q(4,4) then PQ is equal to

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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1. If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point $P$ on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at $P$ passes through $Q (4, 4)$ then $P Q$ is equal to