$P$ is a moving point on the circle $x^2+y^2=4$. $Q$ divide $PN$ ($N$ is foot of perpendicular) in the ratio $1:2$, then locus of $Q$ is

Question

$P$ is a moving point on the circle $x^2+y^2=4$. If a point $Q$ divide the perpendicular dropped from point $P$ on the line $x+y=0$, such that $Q$ divide $PN$ ($N$ is foot of perpendicular) in the ratio $1:2$, then locus of $Q$ is

(a) An ellipse

(b) A Hyperbola

(c) A conic with eccentricity $\dfrac{\sqrt5}{3}$

(d) A Conic with eccentricity $\dfrac{3}{\sqrt5}$

My Solution:

Let a variable point $P$ on circle $P(2\cos(\theta), 2\sin(\theta))$

Foot of perpendicular $N$ from $P$ on Line $x+y=0$ is $N(\cos\theta-\sin\theta, \sin\theta-\cos\theta)$

Let coordinate of $Q$ as $(h,k)$ So according to question $3h=5\cos\theta-\sin\theta .....(i)$ and $3k=5\sin\theta-\cos\theta .....(ii)$

TO find Locus of $Q$ we must remove paramter $\theta$ and this can be done using $(i)^2+(ii)^2$ Hence $(3h)^2+(3k)^2=(5\sin\theta-\cos\theta)^2+(5\cos\theta-\sin\theta)^2$

So Locus of $Q$ using section formula is $x^2+y^2=\dfrac{26}{9}$

What Am I doing wrong?

Answer 1

The parametric equation of $P$ is

$P = (2 \cos t , 2 \sin t ) $

The projection onto the line $x+y= 0$ is obtained as follows

The unit direction vector of this line is $ v = \dfrac{1}{\sqrt{2}}( -1 , 1 )$

And the line passes through the origin. Then the projection is given by

$ N = ( Q \cdot v ) v = ( \sin t - \cos t ) ( -1, 1 ) = (\cos t - \sin t , \sin t - \cos t ) $

Now $Q$ divide $PN$ in the ratio $1:2$, therefore

$ Q = P + \dfrac{1}{3} PN = (2 \cos t, 2 \sin t ) + \dfrac{1}{3} (- \cos t - \sin t , - \cos t - \sin t ) = ( \dfrac{5}{3} \cos t - \dfrac{1}{3} \sin t , \dfrac{5}{3} \sin t - \dfrac{1}{3} \cos t ) $

Grouping the $\cos $ terms and $\sin $ terms together, this becomes

$ Q = \cos t \ ( \dfrac{5}{3} , - \dfrac{1}{3} ) + \sin t \ ( - \dfrac{1}{3} , \dfrac{5}{3} ) $

It is well known that this parametric equation represents an ellipse.

To find the ellipse parameters, there are two methods:

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Method 1:

Replace $t$ in the expression for $Q$ with $ t - \tau $ and let $v_1 = ( \dfrac{5}{3} , - \dfrac{1}{3} ) $ and $v_2 = ( - \dfrac{1}{3} , \dfrac{5}{3} )$, then

$ Q( t , \tau ) = \cos(t - \tau) v_1 + \sin(t - \tau) v_2 $

Expanding the sinusoids, we get

$ Q(t, \tau) = \cos t ( v_1 \cos \tau - v_2 \sin \tau ) + \sin t (v_1 \sin \tau + v_2 \cos \tau ) $

Let

$ w_1( \tau ) = v_1 \cos \tau - v_2 \sin \tau $

$ w_2 (\tau) = v_1 \sin \tau + v_2 \cos \tau $

Then

$ Q(t) = \cos t \ w_1 + \sin t \ w_2 $

We want to choose $\tau$ such that $w_1 \perp w_2$. Performing the dot product between $w_1 $ and $w_2$ we arrive at

$ w_1 \cdot w_2 = (v_1 \cdot v_2) (\cos^2 \tau - \sin^2 \tau) + \cos \tau \sin \tau ( v_1 \cdot v_1 - v_2 \cdot v_2 ) $

Using the double angle formulas, this becomes

$ w_1 \cdot w_2 = (v_1 \cdot v_2) \ \cos(2 \tau) + \frac{1}{2} \sin(2 \tau) ( v_1 \cdot v_1 - v_2 \cdot v2) $

To make this quantity zero, we must have

$ \tan(2 \tau) = \dfrac{\sin(2 \tau) }{\cos(2 \tau) } = \dfrac{ 2 v_1 \cdot v_2 }{ v_2 \cdot v_2 - v_1 \cdot v_1 } $

i.e.

$ \tau = \frac{1}{2} \tan^{-1} \left( \dfrac{2 v_1 \cdot v_2 }{ v_2 \cdot v_2 - v_1 \cdot v_1 } \right) $

With this choice of $\tau$, the vectors $w_1$ and $w_2$ correspond to the semi-major and semi-minor axes vectors (not necessarily in this order).

Let's compute $\tau$ for our ellipse. In this particular case we have

$ v_2 \cdot v_2 - v_1 \cdot v_1 = 0 $ and $ 2 v_1 \cdot v_2 \lt 0 $

Therefore, we should choose $\tau $ as follows

$ \tau = \frac{1}{2} \left( -\dfrac{\pi}{2} \right) = - \dfrac{\pi}{4} $

With this, we can now evaluate $w_1$ and $w_2$

$ w_1 = \dfrac{1}{\sqrt{2}} ( v_1 + v_2 ) = \dfrac{1}{\sqrt{2}} \left( \frac{4}{3} \right) (1, 1) $

And

$ w_2 = \dfrac{1}{\sqrt{2}} (- v_1 + v_2 ) = \dfrac{1}{\sqrt{2}} (2) (-1, 1) $

Hence, the semi-major axis length is

$ a = 2 \times \dfrac{1}{\sqrt{2}} \times \sqrt{(-1)^2 + 1^2} = 2 $

And the semi-minor axis length is

$ b = \dfrac{4}{3} \times \dfrac{1}{\sqrt{2}} \times \sqrt{1^2 + 1^2} = \dfrac{4}{3} $

What remains is to find the eccentricity of the ellipse, based on $a$ and $b$. This is presented below after discussing the second method of finding the ellipse parameters which uses matrix methods.

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Method 2:

In matrix-vector format

$ Q = A u $

where $A = \dfrac{1}{3} \begin{bmatrix} 5 && - 1 \\ -1 && 5 \end{bmatrix} $

and $ u = [ \cos t , \sin t ] $

Since $u^T u = 1 $ , and $ u = A^{-1} Q $, then

$ Q^T A^{-T} A^{-1} Q = 1 $

But,

$ A^{-T} A^{-1} = \left( A A^T \right)^{-1} $

And

$ {A A}^T = \dfrac{1}{9} \begin{bmatrix} 26 && -10 \\ -10 && 26 \end{bmatrix}$

So,

$ \left( {AA}^T \right)^{-1} = \dfrac{9}{ 26^2 - 100} \begin{bmatrix} 26 && 10 \\ 10 && 26 \end{bmatrix} = \dfrac{1}{32} \begin{bmatrix} 13 && 5 \\ 5 && 13 \end{bmatrix} $

The eigenvalues of this matrix are $\dfrac{1}{32}$ of the roots of

$ (\lambda' - 13)^2 - 25 = 0 $

So the eigenvalues are $\lambda_1 = \dfrac{9}{16} $ and $\lambda_2 = \dfrac{1}{4} $

Therefore, the semi-major axis length of the resulting ellipse is $a = 2$ and the semi-minor axis length is $b = \dfrac{4}{3}$.

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The eccentricity is

$e = \sqrt{1 - \left( \dfrac{b}{a} \right)^2 } = \sqrt{ 1 - \left( \dfrac{2}{3} \right)^2 } = \dfrac{\sqrt{5}}{3} $

Therefore, the correct choice is $(c)$.

Answer 2

Another approach is to rotate the coordinate system by 45 degrees clockwise, i.e., $(x',y')=((x-y)/\sqrt{2},(x+y)/\sqrt{2})$. Then the point $P=(x',y')$ on the circle $x'^2+y'^2=4$ projects onto the line $y'=0$ as $N=(x',0)$, so the point of interest is $Q=(x',(2/3)y')$. But this point satisfies $$x'(Q)^2+(3y'(Q)/2)^2=4\implies \frac{x'(Q)^2}{2^2}+\frac{y'(Q)^2}{(4/3)^2}=1$$

So in the rotated coordinate system we have an ellipse with semi-major/minor axes as $a=2,b=4/3$ and thus eccentricity $e=\sqrt{1-b^2/a^2}=\sqrt{5}/3$.