Let A,B and C be the vertices of a variable right angled triangle inscribed in the parabol…

Question

Let $A,B$ and C be the vertices of a variable right angled triangle inscribed in the parabola ${y}^{2}=16x$. Let the vertex $B$ containing the right angle be $(4,8)$ and the locus of the centroid of $	riangleABC$ be a conic ${C}_{0}$. Then three times the length of latus rectum of ${C}_{0}$ is $____$

Answer: 16

Accepted Answer

Since the right angle is at $B=(4,8)$, and $A,C$ lie on the parabola

${y}^{2}=16x,$

we first parametrize the parabola in the standard NCERT form.

For ${y}^{2}=4ax$, we have $4a=16\Rightarrow a=4$.

So any point on the parabola can be written as

$(a{t}^{2},\, 2at)=(4{t}^{2},\, 8t).$

Thus let

$A=(4{t}_{2}^{1},\, 8{t}_{1}),\, C=(4{t}_{2}^{2},\, 8{t}_{2}).$

Also, the point $B=(4,8)$ lies on the parabola and corresponds to parameter $t=1$, since

$(4\cdot {1}^{2},\, 8\cdot 1)=(4,8).$

So $B$ is the point with parameter $1$.

Now, for the triangle $ABC$ to be right-angled at $B$, the slopes of $BA$ and $BC$ must satisfy

${m}_{BA}\, {m}_{BC}=-1.$

For parabola ${y}^{2}=4ax$, the slope of chord joining parameter points ${t}_{1}$ and ${t}_{2}$ is

$\frac{2a({t}_{2}-{t}_{1})}{a({t}_{2}^{2}-{t}_{2}^{1})}=\frac{2}{{t}_{1}+{t}_{2}}.$

So slope of chord joining parameter $1$ and ${t}_{1}$ is

${m}_{BA}=\frac{2}{1+{t}_{1}},$

and slope of chord joining parameter $1$ and ${t}_{2}$ is

${m}_{BC}=\frac{2}{1+{t}_{2}}.$

Since these are perpendicular,

$\frac{2}{1+{t}_{1}}\cdot \frac{2}{1+{t}_{2}}=-1.$

Hence

$\frac{4}{(1+{t}_{1})(1+{t}_{2})}=-1$

which gives

$(1+{t}_{1})(1+{t}_{2})=-4.$

So,

${t}_{1}{t}_{2}+{t}_{1}+{t}_{2}+1=-4$

or

${t}_{1}{t}_{2}+{t}_{1}+{t}_{2}=-5.$

Now let the centroid of $	riangleABC$ be $G(h,k)$.

Using centroid formula,

$h=\frac{4{t}_{2}^{1}+4+4{t}_{2}^{2}}{3}=\frac{4}{3}({t}_{2}^{1}+{t}_{2}^{2}+1),$

$k=\frac{8{t}_{1}+8+8{t}_{2}}{3}=\frac{8}{3}({t}_{1}+{t}_{2}+1).$

Let

$S={t}_{1}+{t}_{2},\, P={t}_{1}{t}_{2}.$

Then from the perpendicularity condition,

$P+S=-5.$

Also,

${t}_{2}^{1}+{t}_{2}^{2}={S}^{2}-2P.$

So

$h=\frac{4}{3}({S}^{2}-2P+1).$

Since $P=-5-S$,

$h=\frac{4}{3}({S}^{2}-2(-5-S)+1)$

$h=\frac{4}{3}({S}^{2}+2S+11).$

Also,

$k=\frac{8}{3}(S+1).$

From this,

$S+1=\frac{3k}{8}\, \Rightarrow \, S=\frac{3k}{8}-1.$

Substitute into $h$:

$h=\frac{4}{3}[{(\frac{3k}{8}-1)}^{2}+2(\frac{3k}{8}-1)+11].$

Now simplify:

${(\frac{3k}{8}-1)}^{2}=\frac{9{k}^{2}}{64}-\frac{3k}{4}+1$

and

$2(\frac{3k}{8}-1)=\frac{3k}{4}-2.$

So inside the bracket,

$\frac{9{k}^{2}}{64}-\frac{3k}{4}+1+\frac{3k}{4}-2+11=\frac{9{k}^{2}}{64}+10.$

Hence

$h=\frac{4}{3}(\frac{9{k}^{2}}{64}+10)=\frac{3{k}^{2}}{16}+\frac{40}{3}.$

Therefore,

$h-\frac{40}{3}=\frac{3{k}^{2}}{16}.$

Rearranging,

${k}^{2}=\frac{16}{3}(h-\frac{40}{3}).$

So the locus is a parabola:

${k}^{2}=4a(h-\frac{40}{3})$

with

$4a=\frac{16}{3}\, \Rightarrow \, a=\frac{4}{3}.$

Length of latus rectum of this parabola is

$4a=\frac{16}{3}.$

Therefore, three times the length of latus rectum is

$3	imes \frac{16}{3}=16.$

So the required answer is

$16.$

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