# Find the equations of the tangent and normal to the hyperbola x²/a² - y²/b² = 1 at the point (x_{0}, y_{0})

**Solution:**

Differentiating x^{2}/a^{2} - y^{2}/b^{2} = 1 with respect to x we have:

2x/a^{2} - 2y/b^{2} dy = 0

⇒ 2y/b^{2} dy/dx = 2x/a^{2}

⇒ dy/dx = b^{2}x/a^{2}y

Therefore, the slope of the tangent at (x_{0}, y_{0}) is

dy/dx](x_{0}, y_{0}) = b^{2}x_{0}/a²y_{0}

Hence, the equation of tangent at (x_{0}, y_{0}) is

y - y_{0} = b^{2}x_{0 }/ a^{2}y_{0} (x - x_{0})

⇒ a^{2} yy_{0} - a^{2} y_{0}^{2} = b^{2} xx_{0} - b^{2} x_{0}^{2}

⇒ b^{2} xx_{0} - a^{2} yy_{0} - b^{2}x_{0}^{2} - a^{2} y_{0}^{2} = 0

⇒ xx_{0 }/ a^{2} - yy_{0}/b^{2} - (x_{0}^{2}/a^{2} - y_{0}^{2}/b^{2}) = 0

⇒ xx_{0 }/ a^{2} - yy_{0}/b^{2} - 1 = 0

⇒ xx_{0 }/ a^{2} - yy_{0 }/ b^{2} = 1

Now, the slope of normal at (x_{0}, y_{0}) is

- 1/slope of the tangent at (x_{0}, y_{0})

= - a^{2}y_{0 }/ b^{2}x_{0}

Hence, the equation of normal at (x_{0}, y_{0}) is

⇒ y - y_{0} = - a^{2}y_{0 }/ b^{2}x_{0} (x - x_{0})

⇒ (y - y_{0}) / a^{2}y_{0 }= - (x - x_{0})/b^{2}x_{0}

⇒ (y - y_{0}) / a^{2}y_{0} + (x - x_{0}) / b^{2}x_{0} = 0

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 24

## Find the equations of the tangent and normal to the hyperbola x²/a² - y²/b² = 1 at the point (x_{0}, y_{0}).

**Summary:**

The equation of the tangent and normal to the hyperbola x^{2}/a^{2} - y^{2}/b^{2} = 1 at the point (x_{0}, y_{0}) is (y - y_{0}) / a^{2}y_{0} + (x - x_{0}) / b^{2}x_{0} = 0

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