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Hyperbola

Section: Mathematics

Question : 9 of 15

Marks: +1, -0

\frac{x^2}{9} - \frac{y^2}{4}

Solution:

Slope of tangent = 2.

The tangents are y =

2x \pm \sqrt{9 \times 4 - 4}

That is, 2x - y =

\pm 4\sqrt{2}

⇒

\frac{x}{2\sqrt{2}} - \frac{y}{4\sqrt{2}}

= 1 and

-\frac{x}{2\sqrt{2}} + \frac{y}{4\sqrt{2}}

= 1

Comparing it with

\frac{xx_1}{9} - \frac{yy_1}{4}

= 1, we get the point of contact as

\left(\frac{9}{2\sqrt{2}} , \frac{1}{\sqrt{2}}\right)

and

\left(-\frac{9}{2\sqrt{2}} , -\frac{1}{\sqrt{2}}\right)

Alternatslope Solution: Equation of tangent at P(θ) is

\left(\frac{\sec\theta}{3}\right) \times -\left(\frac{\tan\theta}{2}\right) y

= 1

⇒ Slope =

\frac{2\sec\theta}{3\tan\theta}

= 2

⇒ sin θ =

\frac{1}{3}

⇒ points are

\left(\frac{9}{2\sqrt{2}},\frac{1}{\sqrt{2}}\right)

and

\left(-\frac{9}{2\sqrt{2}} , -\frac{1}{\sqrt{2}}\right)

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