Question

Find the equation of the tangent to x²/a² - y²/b²=1 at(x₁,y₁).

Answer 1

equation of curve is, x²/a² - y²/b² = 1

differentiate with respect to x,

2x/a² - 2y/b² dy/dx = 0

⇒x/a² = y/b² dy/dx

⇒dy/dx = b²x/a²y

so, slope of tangent of the curve at (x₁,y₁) is m = $\frac{dy}{dx}|_{(x_1,y_1)}$ = $\frac{b^2x_1}{a^2y_1}$

now, equation of tangent is ...

(y - y₁) = m(x - x₁)

⇒y - y₁ = b²x₁/a²y₁ (x - x₁)

⇒a²y₁(y - y₁) = b²x₁(x - x₁ )

⇒a²yy₁ - a²y₁² = b²xx₁ - b²x₁²

⇒ a²yy₁ - b²xx₁ + (b²x₁² - a²y₁²) = 0 This is the equation of tangent of the curve x²/a² - y²/b² at (x₁,y₁).

Answer 2

Answer:

equation of curve is, x²/a² - y²/b² = 1

differentiate with respect to x,

2x/a² - 2y/b² dy/dx = 0

⇒x/a² = y/b² dy/dx

⇒dy/dx = b²x/a²y

so, slope of tangent of the curve at (x₁,y₁) is m = \frac{dy}{dx}|_{(x_1,y_1)}

dx

dy

∣

(x

1

,y

1

)

= \frac{b^2x_1}{a^2y_1}

a

2

y

1

b

2

x

1

now, equation of tangent is ...

(y - y₁) = m(x - x₁)

⇒y - y₁ = b²x₁/a²y₁ (x - x₁)

⇒a²y₁(y - y₁) = b²x₁(x - x₁ )

⇒a²yy₁ - a²y₁² = b²xx₁ - b²x₁²