Question

Two of the normals drawn from a point O to the curve make complementary angles with the axis; prove that the locus of O and the curve which is touched by its polar are parabolas such that their latera recta and that of the original parabola form a geometrical progression. Sketch the three curves.

Answer 1

Let the eqaution of the parabola be $y^2=4ax$ as the normal $y=mx-2am-a{m}^3$ passes through $O$. whose co-ordinates are

$(h,k)$, we have $m_1+m_2+m_3=0$, ...(1)

$\sum m_1{m}_2=\frac{ah-h}{a}$

and $\sum m_1{m}_2{m}_3=-\frac{k}{a}$ ...(3)

Let the angles which the mormals having slopes $m_1$ and $m_2$ make with x-axis be $\theta$ and $\varphi$: then $\theta +\varphi ={90}^o$ or $\varphi =({90}^o-\theta )$

So $m_1=\tan {\theta }_1$ and $m_2=\tan ({90}^o-\theta )=\cot \theta$

$\therefore m_1{m}_2=1$ ...(4)

again $m_3(m_2+m_1)+m_1{m}_2=\frac{2a-h}{a}$ [by (2)]

But $m_1+m_2=-m_3$ [by (1)]

So $-m_3^2+1=\frac{2a-h}{a}$ or $-\frac{k^2}{a^2}=\frac{2a-h}{a}-1=\frac{a-h}{a}$

$\Rightarrow k^2=a(h-a)$. ...(5)

Generalising, we obtain the locus of $O$ as $y^2=a(x-a)$ which as clearly a parabola having latus rectum equal to $a$.

again the equation of the polar of $(h,k)$ w.t., $y^2=4ax$ is

given by $yk=2a(x+h)=2ax+\left(\frac{k^2+a^2}{a}\right)$ from (5)

$\Rightarrow y=\frac{2a}{k}(x+a)+2k$

$\Rightarrow y=\frac{2a}{k}(x+a)+\frac{4a}{2a/k}$.

Changing the oeigin to $(-a,0)$ this becomes $y=\frac{2a}{k}x+\frac{4a}{2a/k}$ which is of the form $y=mx+4a/m$, and hence is tangen to $y^2=4.4ax$

Changimg the origin back to the original one, parabola becomes

$y^2=16a(x+a)$.

Hence the polar of $(h,k)$ntouches a parabola $y^2=16a(x+a)$.

Latus rectum of this parabola is $16a$.

clearly $a,4a,16a$ are in G.P.