Q.24. Let A, B and C be the vertices of a variable right angled triangle inscribed in the parabola y

Q: What is the correct answer for JEE Main Mathematics Q24?

The correct answer is 16. Sol. [Figure: A parabola $y^2=16x$ opening to the right. Point B(4,8) is on the parabola. Two other points A and C are shown on the parabola, such that $\angle ABC = 90^\circ$. A is $(4t_1^2, 8t_1)$ and C is $(4t_2^2, 8t_2)$.] Variable point A & C are shown in the figure :: $m_{AB} \cdot m_{BC} = -1$ $\Rightarrow \frac{8t_1-8}{4t_1^2-4} \cdot \frac{8t_2-8}{4t_2^2-4} = -1$ $\Rightarrow \frac{2(t_1

Q: Q.24. Let A, B and C be the vertices of a variable right angled triangle inscribed in the parabola y = 16x. Let the vertex B containing the right angle be (4, 8) and the locus of the centroid of ABC…

The correct answer is 16. Sol. [Figure: A parabola $y^2=16x$ opening to the right. Point B(4,8) is on the parabola. Two other points A and C are shown on the parabola, such that $\angle ABC = 90^\circ$. A is $(4t_1^2, 8t_1)$ and C is $(4t_2^2, 8t_2)$.] Variable point A & C are shown in the figure :: $m_{AB} \cdot m_{BC} = -1$ $\Rightarrow \frac{8t_1-8}{4t_1^2-4} \cdot \frac{8t_2-8}{4t_2^2-4} = -1$ $\Rightarrow \frac{2(t_1

Question

Q.24. Let A, B and C be the vertices of a variable right angled triangle inscribed in the parabola y = 16x. Let the vertex B containing the right angle be (4, 8) and the locus of the centroid of ABC be a conic C. Then three times the length of latus rectum of C is ________.

FirstInTest · Accepted Answer

Sol. [Figure: A parabola $y^2=16x$ opening to the right. Point B(4,8) is on the parabola. Two other points A and C are shown on the parabola, such that $\angle ABC = 90^\circ$. A is $(4t_1^2, 8t_1)$ and C is $(4t_2^2, 8t_2)$.]

Variable point A & C are shown in the figure
:: $m_{AB} \cdot m_{BC} = -1$
$\Rightarrow \frac{8t_1-8}{4t_1^2-4} \cdot \frac{8t_2-8}{4t_2^2-4} = -1$
$\Rightarrow \frac{2(t_1-1)}{(t_1-1)(t_1+1)} \cdot \frac{2(t_2-1)}{(t_2-1)(t_2+1)} = -1$
$\Rightarrow \frac{4}{(t_1+1)(t_2+1)} = -1$
$\Rightarrow 4 = -(t_1t_2+t_1+t_2+1)$
$\Rightarrow t_1t_2+t_1+t_2 = -5$ ...(1)

Suppose locus of centroid of $	riangle$ABC is (h,k)
h = $\frac{4t_1^2+4t_2^2+4}{3}$, k = $\frac{8t_1+8t_2+8}{3}$
From (1), $t_1+t_2 = -5-t_1t_2$
$t_1t_2 = \frac{k-8}{8/3} = \frac{3k-24}{8}$
$t_1+t_2 = \frac{3h-4-4(t_1^2+t_2^2)}{4}$ (This step is incorrect. It should be $t_1+t_2 = \frac{3k-8}{8}$)

Let's re-derive the centroid coordinates and relation.
Centroid G(h,k) = $(\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3})$
h = $\frac{4t_1^2+4+4t_2^2}{3} = \frac{4(t_1^2+t_2^2+1)}{3}$
k = $\frac{8t_1+8+8t_2}{3} = \frac{8(t_1+t_2+1)}{3}$
From (1), $t_1t_2 = -5-(t_1+t_2)$.
From k, $t_1+t_2 = \frac{3k-8}{8}$.
Substitute into (1): $t_1t_2 = -5 - \frac{3k-8}{8} = \frac{-40-3k+8}{8} = \frac{-3k-32}{8}$.
Now, $t_1^2+t_2^2 = (t_1+t_2)^2 - 2t_1t_2 = (\frac{3k-8}{8})^2 - 2(\frac{-3k-32}{8})$
$= \frac{9k^2-48k+64}{64} + \frac{3k+32}{4} = \frac{9k^2-48k+64+4(3k+32)}{64}$
$= \frac{9k^2-48k+64+12k+128}{64} = \frac{9k^2-36k+192}{64}$.
Substitute into h:
h = $\frac{4}{3} (\frac{9k^2-36k+192}{64} + 1) = \frac{4}{3} (\frac{9k^2-36k+192+64}{64}) = \frac{4}{3} (\frac{9k^2-36k+256}{64})$
h = $\frac{9k^2-36k+256}{48}$
$48h = 9k^2-36k+256$
$9k^2-36k = 48h-256$
$9(k^2-4k) = 48h-256$
$9(k-2)^2 - 36 = 48h-256$
$9(k-2)^2 = 48h-220$
$(k-2)^2 = \frac{48}{9}h - \frac{220}{9} = \frac{16}{3}h - \frac{220}{9}$
This is a parabola of the form $(y-y_0)^2 = 4a(x-x_0)$.
Latus rectum $4a = \frac{16}{3}$.
Length of latus rectum = $\frac{16}{3}$.
Three times the length of latus rectum = $3 	imes \frac{16}{3} = 16$.

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