SATHEE: Hyperbola 3 Question 3

Hyperbola 3 Question 3

3.

Tangents are drawn from any point on the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ to the circle $x^{2} + y^{2} = 9$ . Find the locus of mid-point of the chord of contact.

(2005, 4M)

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Answer:

Correct Answer: 3. $\frac{x^{2}}{9} - \frac{y^{2}}{4} = \frac{{(x^{2} + y^{2})}^{2}}{81}$

Solution:

Let any point on the hyperbola is $(3 \sec θ, 2 \tan θ)$ .

$∴$ Chord of contact of the circle $x^{2} + y^{2} = 9$ with respect to the point $(3 \sec θ, 2 \tan θ)$ is,

$(3 \sec θ) x + (2 \tan θ) y = 9$

Let $(x_{1}, y_{1})$ be the mid-point of the chord of contact.

$\Rightarrow$ Equation of chord in mid-point form is

$x x_{1} + y y_{1} = x_{1}^{2} + y_{1}^{2}$

Since, Eqs. (i) and (ii) are identically equal.

$∴ \frac{3 \sec θ}{x_{1}} = \frac{2 \tan θ}{y_{1}}$

$= \frac{9}{x_{1}^{2} + y_{1}^{2}}$

$\Rightarrow \sec θ = \frac{9 x_{1}}{3 (x_{1}^{2} + y_{1}^{2})}$

$and \tan θ = \frac{9 y_{1}}{2 (x_{1}^{2} + y_{1}^{2})}$

Thus, eliminating ’ $θ$ ’ from above equation, we get

$\frac{81 x_{1}^{2}}{9 {(x_{1}^{2} + y_{1}^{2})}^{2}} - \frac{81 y_{1}^{2}}{4 {(x_{1}^{2} + y_{1}^{2})}^{2}} = 1$

$[∵ \sec^{2} θ - \tan^{2} θ = 1]$

$∴$ Required locus is $\frac{x^{2}}{9} - \frac{y^{2}}{4} = \frac{{(x^{2} + y^{2})}^{2}}{81}$ .

Hyperbola 3 Question 3

3.

Answer:

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Hyperbola 3 Question 3

3.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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