Ellipse and the Right Angle

In this problem, the path of $P$ is an orthoptic , the set of points for which two tangents of a given curve meet at a right angle, and the orthoptic for an ellipse in the form of $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is a circle with the equation $x^2 + y^2 = a^2 + b^2$ . In this problem, since the ellipse has an equation of $\frac{x^2}{16} + \frac{y^2}{9} = 1$ , $a = 4$ and $b = 3$ , so the path of P follows a circle with the equation $x^2 + y^2 = 4^2 + 3^2$ or $x^2 + y^2 = 25$ , which has a radius $r = \sqrt{25} = 5$ and a circumference of $C = 2 \pi r = 2 \pi 5 = 10 \pi$ . Therefore, $k = \boxed{10}$ .

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We can also prove this without knowing the orthoptic shortcut. Using implicit differentiation on $\frac{x^2}{16} + \frac{y^2}{9} = 1$ gives $\frac{x}{8} + \frac{2y}{9}y' = 0$ , which simplifies to $y' = -\frac{9x}{16y}$ , and is the slope of the curve $m = y'$ at any point $(x, y)$ . Let $(p, q)$ and $(r, s)$ be points on the ellipse such that their tangent are perpendicular. Then the equation of the tangent line at a point $(p, q)$ is $y - q = -\frac{9p}{16q}(x - p)$ , and the equation of the tangent line at a point $(r, s)$ is $y - s = -\frac{9r}{16s}(x - r)$ . Now, if $(p, q)$ and $(r, s)$ have tangents that are perpendicular to each other, $m_1m_2 = -1$ , or $(-\frac{9p}{16q})(-\frac{9r}{16s}) = -1$ . Using this and the fact that $(r, s)$ is on the ellipse so $\frac{r^2}{16} + \frac{s^2}{9} = 1$ gives $r = \pm \frac{16^2q}{\sqrt{9^3p^2 + 16^3q^2}}$ and $s = \mp \frac{9^2p}{\sqrt{9^3p^2 + 16^3q^2}}$ . Solving the two tangent line equations for $x$ and $y$ gives $x = \frac{\pm 16q\sqrt{9^3p^2 + 16^3q^2} + 9^2 16p}{9^2p^2 + 16^2q^2}$ and $y = \frac{\mp 9p\sqrt{9^3p^2 + 16^3q^2} + 16^2 9q}{9^2p^2 + 16^2q^2}$ , and $x^2 + y^2$ simplifies (with a lot of effort) to $25$ . Once again, we see that the path of P follows a circle with the equation $x^2 + y^2 = 25$ , which has a radius $r = \sqrt{25} = 5$ and a circumference of $C = 2 \pi r = 2 \pi 5 = 10 \pi$ . Therefore, $k = \boxed{10}$ .

Given any slope $s$ of one of the tangents, we can solve for $x$ the following two equations

$\dfrac{3}{4}\sqrt{16-x^2}=sx+a$
$\dfrac{4}{3}\sqrt{9-x^2}=-\dfrac{1}{s}x+b$

From these, we get the conditions for the lines to be tangent to the ellipse

$a=\sqrt{16+9s^2}$
$b=\dfrac{1}{s}\sqrt{9+16s^2}$

From these, we can find the coordinates of the intersection of the two tangent lines

$x=\dfrac{1}{1+s^2} \left( -s\sqrt{16+9s^2} + \sqrt{9+16s^2}\right)$
$y=sx+\sqrt{16+9s^2}$

Then if we find the distance from this point to the origin $\left(0,0\right)$ , we find that

$\sqrt{x^2+y^2}=5$

Hence the locus is a circle of radius $5$ , and the answer is $k=10$

Let me propose a problem.

Q. When the product of the slopes of 2 tangential on the ellipse $\large\frac{x^2}{a^2} +\frac{y^2}{b^2} =1$ from dot $P(p,q)$ outside of the curve is $\alpha,$ equate $α$ using $a, b, p, \quad \text{and} \quad q.$

|Solution|

We can think of the circle $x^2+y^2=a^2$ , made by multiplying $\frac{a}{b}$ on $y$ -coordinate.

Multiply $\frac{a}{b}$ on the $y$ -coordinate of dot $P$ , and define a new dot $P'(p,\frac{a}{b} q).$

The tangential on the circle $x^2+y^2=a^2$ from dot $P’$ is $y=m(x-p)+\frac{a}{b} q.$

The distance from zero point to the dot of contact $d$ is equal to $a$ , so $d=\frac{|-mp+\frac{a}{b} q|}{\sqrt{m^2+1}}=a.$

Simplified, $a^2 m^2+a^2=m^2 p^2-\frac{2ma}{b} pq+\frac{a^2}{b^2} q^2, \quad m^2 (p^2-a^2 )-\frac{2ma}{b} pq+(\frac{a}{b})^2 (q^2-b^2 )=0.$

This is the quadratic equation on $m$ , and the product of every possible $m$ is $\large(\frac{a}{b})^2 (\frac{q^2-b^2}{p^2-a^2}).$

Since we’d multiplied $\frac{a}{b}$ on the $y$ -coordinate, product of the 2 slopes of the tangential on the circle $x^2+y^2=a^2$ from dot $P’$ is $(\frac{a}{b})^2 α.$

Thus, $α=\large\frac{q^2-b^2}{p^2-a^2}.$

|Application|

Since $αp^2-αa^2=q^2-b^2, αp^2-q^2=αa^2-b^2,$ trace of dot $P(p,q)$ is the quadratic curves.

We need only to observe when $\alpha=-1$ .

Since $p^2-a^2=b^2-q^2, p^2+q^2=a^2+b^2,$ the trace of the dot $P$ is a circle of which the center is the dot $O(0,0)$ and the radius is $\sqrt{a^2+b^2}.$

In this case, $a^2+b^2=5,$ and thus the dot $P$ moves along the circle $x^2+y^2=25$ . Therefore, the length of $P$ 's trace is $10\pi$ , and the value of $k$ is $\boxed{10}$ .

basically the locus of point P comes out as x^2+y^2=a^2+b^2 where a and b are length of semi major axis and semi minor axis respectively of giveen ellipse. hence perimeter =2piR where r=(a^2+b^2)^(1/2) which equals 10pi. note here a=4 and b=3

Let $y=kx+n$ be tangent to the given ellipse. The condition for this is $16k^2+9=n^2$ . So $16k^2+9=(y-kx)^2$ . The equation of the tangent orthogonal to the ellipse is $y=-\dfrac xk+m$ , with $\dfrac{16}{k^2}+9=m^2$ . So $\dfrac{16}{k^2}+9=\Big(y+\dfrac xk\Big)^2$ . Eliminating $k$ from two equations, we obtain $x^6+3 x^4 y^2-39 x^4+3 x^2 y^4-50 x^2 y^2+399x^2+y^6-11 y^4-301 y^2-1225=0$ , i.e. $(x^2+y^2-25)((x^2-7)^2+2 x^2 y^2+y^4+14 y^2)=0$ . Considering that $(x^2-7)^2+2 x^2 y^2+y^4+14 y^2>0$ , there follows $x^2+y^2=25$ , which is the circle equation with $r=5$ , and its circumference is $10\pi$ . So the required value is $10$ .

P is roughly extending to 5 on either axis Cause graph is an ellipse there it won't pass x or y axis without the lines loosing contact 5 + 5 is 10

The locus of pt P will be a director circle of the ellipse. Eqn of director circle will be (x^2+y^2)= a^2+b^2. So radius will be √(9+16)=5. So length= 2πr=10π. K=10.

The locus is a circle called the 'director circle' whose equation is represented as x^2 + y^2 = a^2 + b^2. Here a=5 and b=3 so radius comes out to be 5 and circumference 10π

The trajectory denoted by P is a circle with radius 5, thus the circumference is 10 $\pi$ .