Q.12. Let chord PQ of length of the parabola be such that the ordinates of points P and Q are in t

Q: What is the correct answer for JEE Main Mathematics Q12?

The correct answer is A. Sol. The parabola is $y^2 = 12x$. Comparing with $y^2 = 4ax$, we get $4a=12 \Rightarrow a=3$. The focus is $S(a,0) = S(3,0)$. Let the coordinates of P be $(3t_1^2, 6t_1)$ and Q be $(3t_2^2, 6t_2)$. The ordinates are $6t_1$ and $6t_2$. Given that the ordinates are in the ratio 1:2, so $\frac{6t_1}{6t_2} = \frac{1}{2} \Rightarrow t_2 = 2t_1$. Let $t_1 = t$, then $t_2 = 2t$. So, $P(3t^2, 6t)$ and $Q(

Q: Q.12. Let chord PQ of length of the parabola be such that the ordinates of points P and Q are in the ratio 1: 2. If the chord PQ subtends an angle at the focus of the parabola, then is equal to:

The correct answer is A. Sol. The parabola is $y^2 = 12x$. Comparing with $y^2 = 4ax$, we get $4a=12 \Rightarrow a=3$. The focus is $S(a,0) = S(3,0)$. Let the coordinates of P be $(3t_1^2, 6t_1)$ and Q be $(3t_2^2, 6t_2)$. The ordinates are $6t_1$ and $6t_2$. Given that the ordinates are in the ratio 1:2, so $\frac{6t_1}{6t_2} = \frac{1}{2} \Rightarrow t_2 = 2t_1$. Let $t_1 = t$, then $t_2 = 2t$. So, $P(3t^2, 6t)$ and $Q(

Q: What are the options for this JEE Main Mathematics question?

A. $3/5$ · B. $4/5$ · C. $5/13$ · D. $12/13$

Question

Q.12. Let chord PQ of length  of the parabola  be such that the ordinates of points P and Q are in the ratio 1: 2. If the chord PQ subtends an angle  at the focus of the parabola, then  is equal to:

FirstInTest · Accepted Answer

Sol. The parabola is $y^2 = 12x$. Comparing with $y^2 = 4ax$, we get $4a=12 \Rightarrow a=3$. The focus is $S(a,0) = S(3,0)$. Let the coordinates of P be $(3t_1^2, 6t_1)$ and Q be $(3t_2^2, 6t_2)$. The ordinates are $6t_1$ and $6t_2$. Given that the ordinates are in the ratio 1:2, so $\frac{6t_1}{6t_2} = \frac{1}{2} \Rightarrow t_2 = 2t_1$. Let $t_1 = t$, then $t_2 = 2t$. So, $P(3t^2, 6t)$ and $Q(3(2t)^2, 6(2t)) = Q(12t^2, 12t)$. The length of the chord PQ is given as $3\sqrt{13}$. $PQ^2 = (12t^2 - 3t^2)^2 + (12t - 6t)^2 = (9t^2)^2 + (6t)^2 = 81t^4 + 36t^2$. Given $PQ^2 = (3\sqrt{13})^2 = 9 	imes 13 = 117$. So, $81t^4 + 36t^2 = 117$. Divide by 9: $9t^4 + 4t^2 = 13$. Let $u = t^2$. $9u^2 + 4u - 13 = 0$. $(9u+13)(u-1) = 0$. Since $t^2$ must be positive, $u=t^2=1$. So $t = \pm 1$. Let's take $t=1$. Then $P(3,6)$ and $Q(12,12)$. The chord PQ subtends an angle $\alpha$ at the focus $S(3,0)$. The slope of SP is $m_1 = \frac{6-0}{3-3}$, which is undefined, meaning SP is a vertical line (x=3). The slope of SQ is $m_2 = \frac{12-0}{12-3} = \frac{12}{9} = \frac{4}{3}$. The angle $\alpha$ between SP and SQ can be found using $	an 	heta = \left|\frac{m_2-m_1}{1+m_1m_2}ight|$. Since SP is vertical, the angle it makes with the x-axis is $90^\circ$. The angle SQ makes with the x-axis is $	heta_2 = \arctan(\frac{4}{3})$. So $\alpha = 90^\circ - 	heta_2$. $	an \alpha = 	an(90^\circ - 	heta_2) = \cot 	heta_2 = \frac{1}{	an 	heta_2} = \frac{1}{4/3} = \frac{3}{4}$. Since $	an \alpha = \frac{3}{4}$, and $\alpha$ is an angle in a triangle, $\sin \alpha = \frac{3}{5}$. (Using a right triangle with opposite side 3, adjacent side 4, hypotenuse 5).

Answer

$3/5$

Answer

$4/5$

Answer

$5/13$

Answer

$12/13$

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