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Geometry Level 2

x 2 a 2 + y 2 b 2 = 1 \large \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

If the angle between the tangent drawn to the given ellipse at the parametric point P ( α ) P(\alpha) and the normal drawn to the ellipse at P ( β ) P(\beta) is ϕ \phi such that 0 ∘ ≤ ϕ ≤ 9 0 ∘ 0^{\circ}\leq\phi\leq90^{\circ} and sin ⁡ ( α − β ) = 0 \sin(\alpha-\beta)=0 , then find the value of ϕ \phi .

This is an original problem and belongs to the set My Creations


The answer is 90.

Skanda Prasad by Skanda Prasad , 20, India

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1 solution

Skanda Prasad
Nov 11, 2017

Equation of tangent at P ( α ) P(\alpha) is given by

x a cos ⁡ α + y b sin ⁡ β = 1 \large \dfrac{x}{a}\cos\alpha+\dfrac{y}{b}\sin\beta=1

⟹ \implies It's slope is − b a cot ⁡ α -\dfrac{b}{a}\cot\alpha .

Equation of normal at P ( β ) P(\beta) is

a x cos ⁡ β − b y sin ⁡ β = a 2 − b 2 \large \dfrac{ax}{\cos\beta}-\dfrac{by}{\sin\beta}=a^2-b^2

⟹ \implies It's slope is a b tan ⁡ β \dfrac{a}{b}\tan\beta .

Note that sin ⁡ ( α − β ) = 0 \sin(\alpha-\beta)=0

So, sin ⁡ α cos ⁡ β = cos ⁡ α sin ⁡ β \sin\alpha\cos\beta=\cos\alpha\sin\beta

So, tan ⁡ α tan ⁡ β = 1 \dfrac{\tan\alpha}{\tan\beta}=1

Therefore, tan ⁡ ϕ = ∣ a b tan ⁡ β + b a cot ⁡ α ∣ 1 − tan ⁡ β tan ⁡ α \tan\phi=\dfrac{|\dfrac{a}{b}\tan\beta+\dfrac{b}{a}\cot\alpha|}{1-\dfrac{\tan\beta}{\tan\alpha}}

We can see that tan ⁡ ϕ = ∞ \tan\phi=\infty

Hence, ϕ = 9 0 ∘ \boxed{\phi=90^{\circ}}

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