Question #234668

1. (a) 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.
(b) Find the equations of the tangents to the hyperbola x^2 − 9y^2 = 9 that are drawn from the
point (3, 2) and the points of contact. Find the area of the triangle that these tangents
form with their chord of contact

Expert's answer

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS