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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let one end of a focal chord of the parabola $y^2 = 16x$ be $(16, 16)$. If $P(\alpha, \beta)$ divides this focal chord internally in the ratio $5 : 2$, then the minimum value of $\alpha + \beta$ is equal to:
Solution
$y^2 = 16x$
$\therefore$ 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 $\alpha + \beta = 7$
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