Answer to Question #224022 in Analytic Geometry for Bless

Question #224022

Find the Cartesian equation of the curve C traced out by a point whose coordinates, in terms of a parameter θ, are (a cos θ, b sin θ). Obtain the equations of the tangents at θ = θ1 and θ = θ1 + π/2. Find the coordinates of the points of intersection of the two tangents, and deduce the Cartesian equation of its locus.


1
Expert's answer
2021-09-20T16:04:34-0400

Solution:

Given curve constrained by point acos θ , bsin θ

Now x=acosθ;y=bsinθxa=cosθ,yb=sinθxa2+yb2=1x = a cos θ ; y = b sin θ ⇒ x a = cos θ , y b = sin θ ⇒ x a ^2 + y b ^2 = 1 This is equation of ellipse

Now at point θ\theta

(x1,y1)=(acosθ,bsinθ)(x_1,y_1) =(acos \theta, b sin \theta)

Hence the equation of the tangent

acosθ1xa2+bsinθ1yb2=1xcosθ1xa+ysinθ1b=1\frac{acos \theta_1 x}{a^2}+\frac{bsin \theta_1 y}{b^2}=1\\ \frac{xcos \theta_1 x}{a}+\frac{ysin \theta_1}{b}=1


Now the equation of tangent at point θ;θ1+π2\theta ;\theta_1 +\frac{\pi}{2}

(acosθ,bsinθ)=(acos(θ1+π2),bsin(θ1+π2))=(asinθ1,bcosθ1)(a \cos \theta, b \sin \theta) = (a \cos(\theta_1 +\frac{\pi}{2}), b \sin (\theta_1 +\frac{\pi}{2}))\\ = (-a \sin \theta_1, b \cos\theta_1)

Now equation of tangent

xsinθ1a+ycosθ1b=1-\frac{x\sin \theta_1}{a}+\frac{y\cos \theta_1}{b}=1


The intersection point of both tangents

x=(1usinθb)acosθ1sinθ1a(1usinθb)acosθ1+ycosθ1b=1y=b(sinθ1+cosθ1)Similarlyx=a(cosθ1sinθ1)x= (1- \frac{u \sin \theta}{b}) \frac{a}{\cos \theta_1}\\ -\frac{\sin \theta_1}{a}(1- \frac{u \sin \theta}{b}) \frac{a}{\cos \theta_1}+\frac{y \cos \theta _1}{b}=1\\ y= b(\sin \theta_1 +\cos \theta_1)\\ Similarly \\ x= a(\cos \theta_1 -\sin\theta_1)\\

Hence intersection point

x=a(cosθ1sinθ1)y=b(sinθ1+cosθ1)x= a(\cos \theta_1 -\sin\theta_1)\\ y= b(\sin \theta_1 +\cos \theta_1)\\


Locus of intersection point

x2a2+y2b2=2\frac{x^2}{a^2}+\frac{y^2}{b^2}=2

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