Equation of the tangent to y 2-16 x which is perpendicular to 2 x - y -5-0 isA- x-2 y-17-0B- x-2 y-16-0C- x-2 y-19-0D- x-2 y-18-0

The correct option is B x+2y+16=0 y^2=16x 5ex 2x y+5=0 A=4 0ex m=2 perpendicular slope= 1 2 Eqn of tgt is y=m+a m ⇒ y= 1 2x+4 (1 2) ⇒ y= 1 2x 8 ⇒ 2y= x 16 ⇒ ...

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Byju's Answer

Standard XII

Mathematics

Normal of a Curve y = f(x)

Equation of t...

Question

Equation of the tangent to $y^{2} = 16 x$ which is perpendicular to 2x-y+5=0 is

x+2y+16=0

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x+2y-17=0

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x+2y-19=0

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x+2y-18=0

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Solution

The correct option is A
x+2y+16=0

$y^{2} = 16 x 2 x - y + 5 = 0$
A=4 m=2
perpendicular slope=- $\frac{1}{2}$
Eqn of tgt is $y = m + \frac{a}{m}$
$\Rightarrow y = \frac{- 1}{2} x + \frac{4}{- (\frac{1}{2})}$
$\Rightarrow y = \frac{- 1}{2} x - 8 \Rightarrow 2 y = - x - 16$
$\Rightarrow x + 2 y + 16 = 20$

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Equation of the tangent to y2=16x which is perpendicular to 2x-y+5=0 is

The correct option is A x+2y+16=0 y2=16x2x−y+5=0 A=4 m=2 perpendicular slope=-12 Eqn of tgt is y=m+am ⇒y=−12x+4−(12) ⇒y=−12x−8 ⇒ 2y=−x−16 ⇒x+2y+16=20

Equation of the tangent to $y^{2} = 16 x$ which is perpendicular to 2x-y+5=0 is