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Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

The locus of mid. points of the line segments joining (-3, -5) and the points on the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is :

Option 1 -

$36 x^{2} + 16 y^{2} + 108 x + 80 y + 145 = 0$

Option 2 -

$36 x^{2} + 16 y^{2} + 90 x + 56 y + 145 = 0$

Option 3 -

$9 x^{x} + 4 y^{2} + 18 x + 8 y + 145 = 0$

Option 4 -

$36 x^{2} + 16 y^{2} + 72 x + 32 y + 145 = 0$

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

Let point on ellipse (2sinθ, 3cosθ) and the mid point of line segment joining (-3, -5) and

(2 sin θ, 3cosθ) will be (h, k)

$\frac{2 s i n θ - 3}{2} = h \frac{3 c o s θ - 5}{2} = k$

sin² θ + cos² θ = 1

${(\frac{2 h + 3}{2})}^{2} + {(\frac{2 k + 5}{3})}^{2}$ = 1

$36 x^{2} + 16 y^{2} + 108 x + 80 y + 145 = 0$

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...more

Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

α + 1.6 = 4 ⇒ α = 2.4

β + 2.8 = 6 ⇒ β = 3.2

Ellipse passes through (2.4, 3.2)

⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

Also $1 - \frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{144}{25} b^{2} + \frac{256}{25} a^{2} = a^{2} b^{2}$

$\frac{144}{25} + \frac{256}{25} \times \frac{1}{2} = a^{2}$

$\frac{(128 + 144)}{25} = a^{2} \Rightarrow \frac{272}{25} = a^{2}$

-> $b^{2} = \frac{2 \times 272}{25}$

Latus rectum = $\frac{2 b^{2}}{a}$

(Latus rectum)²

$= \frac{4 b^{4}}{a^{2}} = 4 (\frac{b^{2}}{a^{2}}) b^{2} = \frac{8 \times 272 \times 2}{25} = \frac{4352}{25}$

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The locus of the point of intersection of the lines $(\sqrt[]{3}) k x + k y - 4 \sqrt[]{3} = 0 a n d \sqrt[]{3} x - y - 4 (\sqrt[]{3}) k$ = 0 is a conic, whose eccentricity is…………………

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K $= \frac{4 \sqrt[]{3}}{\sqrt[]{3} x + y}$ $\frac{}{}$

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48 = 16 (e2 – 1)

$e = \sqrt[]{4} = 2$

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The locus of mid. points of the line segments joining (-3, -5) and the points on the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is :

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The locus of mid. points of the line segments joining (-3, -5) and the points on the ellipse x 2 4 + y 2 9 = 1 is :

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The locus of mid. points of the line segments joining (-3, -5) and the points on the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is :