SATHEE: Hyperbola 2 Question 12

Hyperbola 2 Question 12

12.

Tangents are drawn to the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ , parallel to the straight line $2 x - y = 1$ . The points of contacts of the tangents on the hyperbola are

(2012)

(a) $\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}$

(b) $- \frac{9}{2 \sqrt{2}}, - \frac{1}{\sqrt{2}}$

(d) $(- 3 \sqrt{3}, 2 \sqrt{2})$

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Answer:

Correct Answer: 12. (a,b)

Solution:

PLAN: Equation of tangent to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $y = m x \pm \sqrt{a^{2} m^{2} - b^{2}}$

Description of Situation If two straight lines

$a_{1} x + b_{1} y + c_{1} = 0$

and $a_{2} x + b_{2} y + c_{2} = 0$ are identical. Then, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Equation of tangent, parallel to $y = 2 x - 1$ is

$\begin{aligned} y = 2 x \pm \sqrt{9 (4) - 4} \\ ∴ & y = 2 x \pm \sqrt{32} \end{aligned}$

The equation of tangent at $(x_{1}, y_{1})$ is

$\frac{x x_{1}}{9} - \frac{y y_{1}}{4} = 1$

From Eqs. (i) and (ii),

$x_{1} = \frac{9}{2 \sqrt{2}}, y_{1} = \frac{1}{\sqrt{2}}$

Hyperbola 2 Question 12

12.

Answer:

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Hyperbola 2 Question 12

12.

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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