Question

Tangents are drawn to y = cos x from the point P(0, 0). Points of contact of these tangents will always lie on

• A. $\frac { 1 } { x ^ { 2 } } = \frac { 1 } { y ^ { 2 } } + 1$
• B. $\frac { 1 } { x ^ { 2 } } = \frac { 1 } { y ^ { 2 } } - 1$
• C. x² + y² = 1
• D. x²− y² = 1

Answer 1

Answer: (B)

$\frac { 1 } { x ^ { 2 } } = \frac { 1 } { y ^ { 2 } } - 1$

Answer 2

For y = cos x,$\frac { d y } { d x }$ = −sinx.

Let the point of contact be (x₀, cosx₀). equation of tangent to y = cosx at this point is (y - cosx₀) = sinx₀(x₀ - x). It must pass through (0, 0).

Thus x₀ = -cotx₀. We also have y₀ = cosx₀

⇒ $\frac { 1 } { y _ { 0 } ^ { 2 } } - \frac { 1 } { x _ { 0 } ^ { 2 } }$. Thus required locus is $\frac { 1 } { y ^ { 2 } } - \frac { 1 } { x ^ { 2 } }$ =1