If the normal at the point $$P($$θ$$)$$ to the ellipse $$x214+y5=1$$ intersects it again at the point $$Q(2$$θ$$)$$, then $$cos$$⁡θ is equal to

Question

If the normal at the point $$P($$θ$$)$$ to the ellipse $$x214+y5=1$$ intersects it again at the point $$Q(2$$θ$$)$$, then $$cos$$⁡θ is equal to

Infinity Learn · Accepted Answer

The normal at $$P(acos$$⁡θ,bsin⁡θ$$)$$ is axcos⁡θ−bxsin⁡θ$$=a2$$−$$b2$$, where $$a2=14$$,$$b2=5$$ It meets the curve again at $$Q(2$$θ$$)$$ , i.e., (acos$$⁡$$2$$θ,bsin⁡$$2$$θ$$)∴ acos⁡θ(acos$$⁡$$2$$θ$$)−bsin⁡θ(bsin$$⁡$$2$$θ$$)=a2$$−$$b2$$⇒ $$14cos$$⁡θ$$($$$$cos$$⁡$$2$$θ$$)−$$5sin$$⁡θ$$($$$$sin$$⁡$$2$$θ$$)=14$$−$$5$$⇒ $$28cos2$$⁡θ−$$14$$−$$10cos2$$⁡θ$$=9cos$$⁡θ⇒ $$18cos2$$⁡θ−$$9cos$$⁡θ−$$14=0$$⇒ $$(6cos$$⁡θ−$$7)(3cos$$⁡θ$$+2)=0$$⇒$$cos$$⁡θ$$=-23$$

$If the normal at the point P (θ) to the ellipse \frac{x^{2}}{14} + \frac{y}{5} = 1 intersects it again at the point Q (2 θ), then \cos θ is equal to$

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If the normal at the point P(θ) to the ellipse x214+y5=1 intersects it again at the point Q(2θ), then cos⁡θ is equal to

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$If the normal at the point P (θ) to the ellipse \frac{x^{2}}{14} + \frac{y}{5} = 1 intersects it again at the point Q (2 θ), then \cos θ is equal to$