Name: Tangents and Normal
Uploaded: 2019-07-19
Duration: 14 min 38 s
Description: In this video, you will learn how to find the slopes of a tangent and normal, and the point of contact. It will help you solve questions in exams.

Question

Find the equation of the normal to the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

at

(a \sec θ, b \tan θ) .

The normals to the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

at the points

(a \sec ϕ, b \tan ϕ)

and

(a \sec θ, b \tan θ),

where,

θ + ϕ = \frac{π}{2},

meet at

(h, k) .

Find

k .

Accepted Answer

The equation of hyperbola: $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

Differentiating both sides w.r.t $x$ , we get: $\frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \times \frac{d y}{d x} = 0$

$\Rightarrow \frac{2 x}{a^{2}} = \frac{2 y}{b^{2}} \times \frac{d y}{d x}$

$\Rightarrow \frac{b^{2} x}{a^{2} y} = \frac{d y}{d x}$

$\Rightarrow {(\frac{d y}{d x})}_{(a \sec θ, b \tan θ)} = \frac{b^{2} a \sec θ}{a^{2} b \tan θ} = \frac{b \sec θ}{a \tan θ}$

The required equation of normal at $(a \sec θ, b \tan θ)$ is

$y - b t an θ = \frac{- 1}{\frac{b \sec θ}{a \tan θ}} (x - a \sec θ)$ [Using $y - y_{1} = \frac{- 1}{{(\frac{d y}{d x})}_{(a \sec θ, b \tan θ)}} (x - x_{1})$ ]

$\Rightarrow y - b \tan θ = \frac{- atan θ}{b \sec θ} (x - a \sec θ)$

$\Rightarrow b \sec θ y - b^{2} \tan θ \sec θ = - a x \tan θ + a^{2} \tan θ \sec θ$

$\Rightarrow a x \tan θ + b \sec θ y = a^{2} \tan θ \sec θ + b^{2} \tan θ \sec θ$

$\Rightarrow a x \tan θ + b \sec θ y = \tan θ \sec θ (a^{2} + b^{2})$

$\Rightarrow a x \cos θ + b y \cot θ = a^{2} + b^{2}$

And similarly, the equation of the normal at $(a \sec ϕ, b \tan ϕ)$ is $a x \cos ϕ + b y \cot ϕ = a^{2} + b^{2}$

It is given that, the normals at the point $(a \sec ϕ, b \tan ϕ) and (a \sec θ, b \tan θ)$ is meets the point at $(h, k)$ i.e., $∴ a h \cos θ + b k \cot θ = a^{2} + b^{2} \to (1)$ and $a h \sin θ + b k \tan θ = a^{2} + b^{2} \to (2)$ [Using $θ + ϕ = \frac{π}{2}$ ]

Now, equation $(1) \times \sin θ -$ equation $(2) \times \cos θ$ , then we get: $(b k + a^{2} + b^{2}) (\sin θ - \cos θ) = 0$

$\Rightarrow b k = - (a^{2} + b^{2})$ [since $(\sin θ - \cos θ) \neq 0$ ]

Thus, the required value of $k = - (\frac{a^{2} + b^{2}}{b})$ .

Important Questions on Tangent and Normal

Learn and Prepare for any exam you want !

Find the equation of the normal to the hyperbola x2a2-y2b2=1 at (asecθ,btanθ). The normals to the hyperbola x2a2-y2b2=1 at the points (asecϕ,btanϕ) and (asecθ,btanθ), where, θ+ϕ=π2, meet at (h,k). Find k.

Important Questions on Tangent and Normal

Learn and Prepare for any exam you want !