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If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is $16x^2+9y^2+k_1x-36y+k_2=0$, then find $k_1$ and $k_2$

Before solving it, I noticed a problem with it. Even if we reflect the curve, the coefficient of $x^2$ is 9, and not 16 according to what the question says.

That’s all I have as doubt. I don’t need the whole answer, just need to know if the question is right or not.

Aditya
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  • not "the curve on the reflection" but 'the curve which is the symmetrical of.." – Jean Marie Mar 21 '20 at 16:27
  • @JeanMarie I too got confused on that part, but that’s exactly how the question is written. – Aditya Mar 21 '20 at 18:09
  • The equation of an ellipse is not unique. The coefficient of $x^2$ in the equation of the reflected ellipse no more “must be” $9$ than the coefficient of $x^2$ in the original equation “must be” $1/16$. That aside, looks like you’re getting numerators and denominators confused. – amd Mar 21 '20 at 21:34
  • @JeanMarie Read “on the reflection of the ellipse...” as “obtained upon reflecting the ellipse...” Not a common construction nowadays, but grammatically and syntactically valid. – amd Mar 21 '20 at 21:35

4 Answers4

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Note that the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ with respect to the line $x-y-2=0$, which has a tangent angle of 45-degrees, is the ellipse that is centered at $(5,2)$, the reflection point of $(4,3)$, and is rotated at 90-degrees, as shown in the graph

enter image description here

So, the equation of the reflection ellipse is,

$$\frac{(x-5)^2}{9}+\frac{(y-2)^2}{16}=1$$

Then, compare with the given equation $16x^2+9y^2+k_1x-36y+k_2=0$ to obtain $k_1=-160$ and $k_2=292$.

Quanto
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  • Is there a reason that the ellipse got rotated 90 degees? It’s obvious from the graph, but looks hard to be done analytically – Aditya Mar 22 '20 at 04:27
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    @Aditya - Because the reflection line is at the 45-degrees with the major axis of the given ellipse, it is then at the 45-degrees with the major axis of the reflection ellipse. Thus, the two major axes are at 90 degrees. – Quanto Mar 22 '20 at 05:03
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Consider the line $y=x-2$;

New coordinates : $X=x$; and $Y=y+2$;

Reflect about $Y=X.$

The given curve expressed in $X,Y:$

$(X-4)^2/4^2+(Y-5)^2/3^2=1$;

The reflected curve:

$(Y-4)^2/4^2+(X-5)^2/3^2=1$;

Back to $x,y$:

$(y-2)^2/4^2+(x-5)^2/3^2=1$;

$9(y-2)^2+16(x-5)^2=16\cdot 9$;

Can you finish?

Peter Szilas
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  • Why can’t we simply find the reflection of the centre of ellipse, and then keep the major and minor axis the same? – Aditya Mar 21 '20 at 18:11
  • Aditya. Look at Quanto"s answer, drawing. The result is an ellipse with major and minor axes interchanged .Easier: Reflect an ellipse on.y=x. Simply interchange x and y in the equation, what happens? Start : you have major axis x : equals 4 , after refection minor axis x :equals 3, similarly for y. – Peter Szilas Mar 21 '20 at 19:15
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$$\frac{(x-4)^2}{16}+\frac{(y-3)}{9}=1 \implies x=4 \cos t+4,y=3 \sin t+3 ~~~(1).$$ The image $(X,Y)$ of $(x,y)$ in the line $ax+by+c=0$ is given by: $$\frac{X-x}{a}=\frac{Y-y}{b}=-2\frac{(ax+by+c)}{a^2+b^2}$$ So we get $$\frac{X-4\cos t-4}{1}=\frac{Y-3\sin t-3}{-1}=-2\frac{4\cos t+4-3\sin t -3-2}{2}$$, we get $$X=-4\cos t-4 +4\cos t+4+3\sin t+3+2=3 \sin t+5$$ $$ Y=4 \cos t+4-2= 4\cos t +2$$ $$\implies \sin t=\frac{X-5}{3}, ~~ \cos t=\frac{Y-2}{4}$$ Squaring and adding these two results we get the required image the ellipse (1) as $$\frac{(X-5)^2}{9}+\frac{(Y-2)^2}{16}=1.$$ Finally, one would write the image ellipse as $$\frac{(x-5)^2}{9}+\frac{(y-2)^2}{16}=1.$$ in the same plane as (1)

Z Ahmed
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  • In the second line, why was -2 multiplied to $ax+by+c$ – Aditya Mar 22 '20 at 04:28
  • It is (-2) if yyou want image of (x.y) in the line. It is (-1) if want foot of perpendicular then use (-1). When you solve the general problem of image/foot on line ax+by+c=0 by ordinary co-ordinate geometry this result emerges. – Z Ahmed Mar 22 '20 at 04:44
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WLOG any point on the ellipse be $P(4+4\cos t, 3+3\sin t )$

If the reflected point is $Q(h,k)$

The midpoint of $PQ$ will lie on $$x-y-2=0$$

As $PQ\perp$ to $x-y-2=0,$ the gradient of $PQ$ will be $$\dfrac{-1}1$$

So, we have two simultaneous equations in $h,k$

Solve them and eliminate $t$ using $$\cos^2t+\sin^2t=1$$

Matteo
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lab bhattacharjee
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