BITSAT2015MathematicsApplication of DerivativesActual
Tangents are drawn from the origin to the curve y = x . Their points of contact lie on
Options
- Ax² y²=y²-x²
- Bx² y²=x²+y²
- Cx² y²=x²-y²
- DNone of these
Correct answer
C. x² y²=x²-y²
Step-by-step solution
Let (x₁, y₁ ) be one of the points of contact. Given curve is y= x array l dy dx =- x . dy dx |_ ( x ₁, y ₁ ) =- x ₁ array Now the equation of the tangent at ( x ₁, y ₁ ) is aligned & y-y₁ ( d y d x )_ (x₁, y₁ ) (x-x₁ ) & y-y₁=- x₁ (0-x₁ ) aligned Since, it is given that equation of tangent passes through origin. array lr & 0-y₁=- x₁ (0-x₁ ) & y₁=-x₁ x₁ array Also, point (x₁, y₁ ) lies on y= x y₁= x₁ From Eqs. (i), (ii), we get aligned & ² x₁+ ² x₁= y₁² x₁² +y₁²=1 & x₁²=y₁²+y₁² x₁² aligned Hence, the locus of (x₁,