SATHEE: Hyperbola 2 Question 6

Hyperbola 2 Question 6

6.

If a hyperbola passes through the point $P (\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$ , then the tangent to this hyperbola at $P$ also passes through the point

(2017 Main)

(a) $(3 \sqrt{2}, 2 \sqrt{3})$

(b) $(2 \sqrt{2}, 3 \sqrt{3})$

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

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Answer:

Correct Answer: 6. (b)

Solution:

Let the equation of hyperbola be $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .

$\begin{array}{lc} ∴ & a e = 2 \Rightarrow a^{2} e^{2} = 4 \\ \Rightarrow & a^{2} + b^{2} = 4 \Rightarrow b^{2} = 4 - a^{2} \\ ∴ & \frac{x^{2}}{a^{2}} - \frac{y^{2}}{4 - a^{2}} = 1 \end{array}$

Since, $(\sqrt{2}, \sqrt{3})$ lie on hyperbola.

$\begin{aligned} ∴ \frac{2}{a^{2}} - \frac{3}{4 - a^{2}} = 1 \\ \Rightarrow 8 - 2 a^{2} - 3 a^{2} = a^{2} (4 - a^{2}) \\ \Rightarrow 8 - 5 a^{2} = 4 a^{2} - a^{4} \\ \Rightarrow a^{4} - 9 a^{2} + 8 = 0 \\ \Rightarrow (a^{4} - 8) (a^{4} - 1) = 0 \Rightarrow a^{4} = 8, a^{4} = 1 \\ ∴ a = 1 \end{aligned}$

Now, equation of hyperbola is $\frac{x^{2}}{1} - \frac{y^{2}}{3} = 1$ .

$∴$ Equation of tangent at $(\sqrt{2}, \sqrt{3})$ is given by

$\sqrt{2} x - \frac{\sqrt{3} y}{3} = 1 \Rightarrow \sqrt{2} x - \frac{y}{\sqrt{3}} = 1$

which passes through the point $(2 \sqrt{2}, 3 \sqrt{3})$ .

Hyperbola 2 Question 6

6.

Answer:

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Hyperbola 2 Question 6

6.

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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