SATHEE: Ellipse 2 Question 11

Ellipse 2 Question 11

11. The normal at a point $P$ on the ellipse $x^{2} + 4 y^{2} = 16$ meets the $X$ -axis at $Q$ . If $M$ is the mid-point of the line segment $P Q$ , then the locus of $M$ intersects the latusrectum of the given ellipse at the points (2009)

(a) $\pm \frac{3 \sqrt{5}}{2}, \pm \frac{2}{7}$

(b) $\pm \frac{3 \sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4}$

(d) $\pm 2 \sqrt{3}, \pm \frac{4 \sqrt{3}}{7}$

Show Answer

Solution:

Given, $\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$

[replacing $k$ by $y$ and $h$ by $x$ ]

Here,

$a = 4, b = 2$

Equation of normal

$\begin{aligned} 4 x \sec θ - 2 y cosec θ & = 12 \\ M \frac{7 \cos θ}{2}, \sin θ & = (h, k) \\ ∴ & h & = \frac{7 \cos θ}{2} \\ \Rightarrow & = \frac{2 h}{7} = \cos θ \end{aligned}$

and $k = \sin θ$

On squaring and adding Eqs. (i) and (ii), we get

$\begin{aligned} \frac{4 h^{2}}{49} + k^{2} & = 1 [∵ \cos^{2} θ + \sin^{2} θ = 1] \\ Hence, locus is \frac{4 x^{2}}{49} + y^{2} & = 1 \end{aligned}$

For given ellipse, $e^{2} = 1 - \frac{4}{16} = \frac{3}{4}$

$\begin{aligned} ∴ e = \frac{\sqrt{3}}{2} \\ ∴ x = \pm 4 \times \frac{\sqrt{3}}{2} = \pm 2 \sqrt{3} [∵ x = \pm a e] \end{aligned}$

On solving Eqs. (iii) and (iv), we get

$\begin{aligned} \frac{4}{49} \times 12 + y^{2} & = 1 \Rightarrow y^{2} = 1 - \frac{48}{49} = \frac{1}{49} \\ y & = \pm \frac{1}{7} \end{aligned}$

$∴$ Required points $\pm 2 \sqrt{3}, \pm \frac{1}{7}$ .

पिछला

अगला

Ellipse 2 Question 11

11. The normal at a point $P$ on the ellipse $x^{2} + 4 y^{2} = 16$ meets the $X$ -axis at $Q$ . If $M$ is the mid-point of the line segment $P Q$ , then the locus of $M$ intersects the latusrectum of the given ellipse at the points (2009)

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

Ellipse 2 Question 11

11. The normal at a point P on the ellipse x2+4y2=16 meets the X-axis at Q. If M is the mid-point of the line segment PQ, then the locus of M intersects the latusrectum of the given ellipse at the points (2009)

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

Menu

11. The normal at a point $P$ on the ellipse $x^{2} + 4 y^{2} = 16$ meets the $X$ -axis at $Q$ . If $M$ is the mid-point of the line segment $P Q$ , then the locus of $M$ intersects the latusrectum of the given ellipse at the points (2009)