The equation of tangent at $(5, 3)$ for the curve $x^{2} - y^{2} = 16$

Open in App

Solution

Given point $(5, 3)$
Given equation of curve $x^{2} - y^{2} = 16$
Slope of tangent is given as $2 x - 2 y \frac{d y}{d x} = 0 \frac{d y}{d x} = \frac{x}{y} {\frac{d y}{d x} ∣ ∣ ∣}_{(5, 3)} = \frac{5}{3}$
Slope of tangent is $(y - 3) = \frac{5}{3} (x - 5) 3 y - 9 = 5 x - 25 5 x - 3 y - 16 = 0$

Suggest Corrections

Similar questions

\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1

y -

\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1

(8, 3 \sqrt{3})

\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1

(8, 3 \sqrt{3})

Find the equation of the tangents drawn form the point (5,3) to the hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{9} = 1.$

Find the equation of the tangents draws form the point (5,3) to the hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{9} = 1.$

Join BYJU'S Learning Program