JEE MAIN Parabola Previous Year Questions (PYQs) – Page 5 of 5

$h = \frac{4t}{5}$

$k = \frac{2t^2}{5} = \frac{2}{5}\left(\frac{5h}{4}\right)^2$

$8k = 5h^2$

$\Rightarrow 5x^2 = 8y$

$T = S_1$

$5(xx_1) - 4(y + y_1) = 5x_1^2 - 8y_1$

$5(xx_1) - 4(y + 2) = 5 - 8 \cdot 2$

$5x - 4y + 3 = 0$

$\therefore$ parameter of point $A$ is $t = 2$

$\Rightarrow$ Parameter of point $B$ is $t = -\frac{1}{2}$

$\Rightarrow$ Coordinates of $B$ is $(1, -4)$

Case 1:

$A(16,16),; P(\alpha,\beta),; B(1,-4)$

$\alpha = \frac{5\cdot1 + 2\cdot16}{7} = \frac{37}{7}$

$\beta = \frac{5(-4) + 2\cdot16}{7} = \frac{12}{7}$

$\Rightarrow \alpha + \beta = 7$

Case 2:

$A(16,16),; P(\alpha,\beta),; B(1,-4)$

$\alpha = \frac{2\cdot1 + 5\cdot16}{7}$

$\beta = \frac{2(-4) + 5\cdot16}{7}$

$\Rightarrow \alpha + \beta = 22$

So minimum value of $\alpha + \beta = 7$