First slide

Parabola

Question

$A \tan g e n t i s d r a w n t o t h e p a r a b o l a$ $y^{2} = 6 x$ $w h i c h i s p e r p e n d i c u l a r t o t h e l i n e$ $2 x + y = 1$ . $W h i c h o f t h e f o l l o w i n g p o i n t s d o e s N O T l i e o n i t$

Moderate

A

$(5, 4)$
B

$(4, 5)$
C

$(0, 3)$
D

$(- 6, 0)$

Solution

The equation of the given parabola is $y^{2} = 6 x$
differentiate both sides
$2 y \frac{d y}{d x} = 6 \Rightarrow \frac{d y}{d x} = \frac{3}{y}$
the slope of the line $2 x + y = 1$ is $- 2$ ,
since the above line is perpendicular to the required tangent
$\frac{3}{y} (- 2) = - 1 y = 6$
when $y = 6$ , then $x = 6$
The equation of tangent is
$y - 6 = \frac{1}{2} (x - 6) 2 y - 12 = x - 6 x - 2 y + 6 = 0$
verify the options

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