Name: Tangent & Normal to Ellipse
Uploaded: 2019-07-19
Duration: 4 min 30 s
Description: We have studied the tangent to a circle. A circle is a special case of ellipses. Learn more about a tangent to an ellipse by watching this video.

Question

Find the equations to the tangents to the ellipse,

x^{2} + 2 y^{2} = 3

drawn from the point

(1, 2)

and also find the angle between these tangents.

Accepted Answer

Equation of the ellipse $x^{2} + 2 y^{2} = 3$

$\Rightarrow \frac{x^{2}}{3} + \frac{y^{2}}{\frac{3}{2}} = 1$

$\Rightarrow a^{2} = 3, b^{2} = \frac{3}{2}$

Let $m$ be the slope of the tangent which is passing through $P (1, 2)$

Equation of the tangent is $y - 2 = m (x - 1) = m x - m$

$y = m x + (2 - m)$

Above line is a tangent to the ellipse

$\Rightarrow c^{2} = a^{2} m^{2} + b^{2}$

$\Rightarrow (2 - m)^{2} = 3 (m^{2}) + \frac{3}{2}$

$\Rightarrow 4 + m^{2} - 4 m = 3 m^{2} + \frac{3}{2}$

$\Rightarrow 2 m^{2} + 4 m - \frac{5}{2} = 0$

$\Rightarrow 4 m^{2} + 8 m - 5 = 0$

$\Rightarrow (2 m - 1) (2 m + 5) = 0$

$m = \frac{1}{2} or - \frac{5}{2}$

Case $(i)$ $m = \frac{1}{2}$

Equation of the tangent is $\Rightarrow y = \frac{1}{2} x + 2 - \frac{1}{2} = \frac{x}{2} + \frac{3}{2}$

$\Rightarrow 2 y = x + 3$

$\Rightarrow x - 2 y + 3 = 0$

Case $(i i)$ $m = - \frac{5}{2}$

Equation of the tangent is $y = - \frac{5}{2} x + (2 + \frac{5}{2}) = - \frac{5 x}{2} + \frac{9}{2}$

$\Rightarrow 2 y = - 5 x + 9$

$\Rightarrow 5 x + 2 y - 9 = 0$

Angle between the tangents is given by $\Rightarrow \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}|$

$= |\frac{\frac{1}{2} + \frac{5}{2}}{1 + (\frac{1}{2}) (- \frac{5}{2})}| = |\frac{3}{1 - \frac{5}{4}}| = | - 12 | = 12$

$\Rightarrow θ = \tan^{- 1} (12)$

Andhra Pradesh Board Solutions for Exercise 2: Exercise 4(b)

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 2: Exercise 4(b)

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