Test Index

BITSAT 2015 Paper

Section: Mathematics

Question : 105 of 150

Marks: +1, -0

Solution:

Let

(x_1, y_1)

be one of the points of contact.

Given curve is y = cos x

⇒

\frac{dy}{dx}

= - sin x

\left.\frac{dy}{dx}\right|_{(x_1, y_1)}

= - sin

x_1

Now the equation of the tangent at

(x_1, y_1)

y -

y_1 \left(\frac{dy}{dx}\right)_{(x_1, y_1)}

(x-x_1)

⇒

y-y_1

= - sin

x_1

(0 -

x_1

)

⇒

y_1

-x_1 \sin x_1

... (i)

Also, point

(x_1, y_1)

lies on y = cos x.

∴

y_1

= cos

x_1

From Eqs. (i), (ii) , we get

\sin^2 x_1 + \cos^2 x_1

\frac{y_1^2}{x_1^2} + y_1^2

= 1

⇒

x_1^2

y_1^2 + y_1^2 x_1^2

Hence, the locus of

(x_1, y_1)

x^2

y^2 + y^2 x^2

⇒

x^2 y^2

x^2 - y^2

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