Question

Q35⟩⟩ Find the equations of tangent and normal to the ellipse x² + 4y² = 32 when θ = π/4.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation of ellipse is

$\rm \: {x}^{2} + {4y}^{2} = 32$

can be further rewritten as

$\rm \: \dfrac{ {x}^{2} }{32} + \dfrac{ {4y}^{2} }{32} = 1$

$\rm \: \dfrac{ {x}^{2} }{32} + \dfrac{ {y}^{2} }{8} = 1 - - - (1)$

On comparing with general equation of ellipse

$\rm \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

we get

$\rm \: {a}^{2} = 32\rm\implies \:a = 4 \sqrt{2}$

and

$\rm \: {b}^{2} = 8 \: \: \rm\implies \:a = 2 \sqrt{2}$

Now, We know

Equation of tangent to the ellipse $\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} }=1$ at $\theta$ is given by

$\boxed{ \rm{ \:\dfrac{x \: cos \theta}{a} + \dfrac{y \: sin \theta}{b} = 1 \: \: }}$

So, equation of tangent to equation (1) at $\theta = \dfrac{\pi}{4}$ is

$\rm \: \dfrac{x \: cos\dfrac{\pi}{4}}{4 \sqrt{2} } + \dfrac{y \: sin\dfrac{\pi}{4}}{2 \sqrt{2} } = 1$

$\rm \: \dfrac{x \:}{4 \sqrt{2} \times \sqrt{2} } + \dfrac{y \: }{2 \sqrt{2} \times \sqrt{2} } = 1$

$\rm \: \dfrac{x \:}{8} + \dfrac{y \: }{4} = 1$

$\rm \: \dfrac{x \: + \: 2y}{8} = 1$

$\rm\implies \:\boxed{ \rm{ \:x + 2y = 8 \: \: }}$

Now, Equation of normal

Equation of normal to the ellipse $\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} }=1$ at $\theta$ is given by

$\rm \: \boxed{ \rm{ \:\dfrac{ax}{cos\theta} - \dfrac{by}{sin\theta} = {a}^{2} - {b}^{2} \: }}$

So, equation of normal to equation (1) at $\theta = \dfrac{\pi}{4}$ is

$\rm \: \dfrac{4 \sqrt{2} x}{cos\dfrac{\pi}{4}} - \dfrac{2 \sqrt{2} y}{sin\dfrac{\pi}{4}} = 32 - 8 \:$

$\rm \: 4 \sqrt{2} \times \sqrt{2}x - 2 \sqrt{2} \times \sqrt{2} y \: = \: 24$

$\rm \: 8x - 4 y \: = \: 24$

$\rm \: 8(2x - y) \: = \: 24$

$\rm\implies \:\boxed{ \bf{ \:2x - y = 3 \: }}$