Write down the equations to the tangent and normal at the point (4, 6) of the parabola $y^{2} = 9 x .$

Open in App

Solution

Equation of tangent at any point $(h, k)$ is $T = 0$

So equation of tangent at $(4, 6)$ of the parabola $y^{2} = 9 x$ is

$6 y = 9 (\frac{x + 4}{2}) 12 y = 9 x + 36 9 x - 12 y + 36 = 0 3 x - 4 y + 12 = 0$

Slope of tangent $= m = - \frac{a}{b} = - \frac{3}{- 4} = \frac{3}{4}$

$\Rightarrow$ Slope of normal $= - \frac{4}{3}$

So the equation of normal at $(4, 6)$ is

$y - 6 = - \frac{4}{3} (x - 4) 3 y - 18 = - 4 x + 16 4 x + 3 y = 34$

Suggest Corrections

Similar questions

y^{2} = 6 x

y^{2} = 12 x

Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at², 2at).

y^{2} = 4 a x

(a t^{2}, 2 a t)

y^{2} = 4 a (x - a) .

Join BYJU'S Learning Program