Answer to Question #224022 in Analytic Geometry for Bless

Question

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.

Answer 1

Solution:

Given curve constrained by point acos θ , bsin θ

Now $x = a cos θ ; y = b sin θ ⇒ x a = cos θ , y b = sin θ ⇒ x a ^2 + y b ^2 = 1$ This is equation of ellipse

Now at point $\theta$

$(x_1,y_1) =(acos \theta, b sin \theta)$

Hence the equation of the tangent

$\frac{acos \theta_1 x}{a^2}+\frac{bsin \theta_1 y}{b^2}=1\\ \frac{xcos \theta_1 x}{a}+\frac{ysin \theta_1}{b}=1$

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

$(a \cos \theta, b \sin \theta) = (a \cos(\theta_1 +\frac{\pi}{2}), b \sin (\theta_1 +\frac{\pi}{2}))\\ = (-a \sin \theta_1, b \cos\theta_1)$

Now equation of tangent

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

The intersection point of both tangents

$x= (1- \frac{u \sin \theta}{b}) \frac{a}{\cos \theta_1}\\ -\frac{\sin \theta_1}{a}(1- \frac{u \sin \theta}{b}) \frac{a}{\cos \theta_1}+\frac{y \cos \theta _1}{b}=1\\ y= b(\sin \theta_1 +\cos \theta_1)\\ Similarly \\ x= a(\cos \theta_1 -\sin\theta_1)\\$

Hence intersection point

$x= a(\cos \theta_1 -\sin\theta_1)\\ y= b(\sin \theta_1 +\cos \theta_1)\\$

Locus of intersection point

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