Question

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Answer 1

Answer:

• Parallel : 2x−y+2=0 ; Point of contact = (1,4)
• Perpendicular : x+2y+16=0 ; Point of contact = (16,-16)

Step-by-step explanation:

Given :

y² = 16x is a parabola

To find :

• Equations of tangents which are parallel and perpendicular to 2x−y+5=0.
• co-ordinates of their points of contact.

Solution :

Equation of the parabola is y² = 16x [a = 4]

The tangent is parallel to 2x−y+5=0

Equation of the tangent can be taken as y=2x+c

We know that, If a line is tangent to a parabola then, c = a/m

c = 4/2 = 2

Thus, Equation of the tangent can be taken as y=2x+2

⇒2x−y+2=0

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We know that, The coordinates of point of contact is given by

$\bigg(\dfrac{a}{m^2} \ , \, \dfrac{2a}{m}\bigg)$

$\text{Point of contact = } \mathrm {\bigg(\dfrac{4}{4}, \dfrac{8}{2} \bigg) } = (1,4)$

__________________________

Perpendicular slope to given line is -1/2 ( = m')

So, tangent formed by it is of the form $y = -\dfrac{1}{2}x + c'$

Now,

$c' = \dfrac{a}{m'} = \dfrac{4}{(-\frac{1}{2} )} = -8$

Equation of the perpendicular tangent is

$y = -\dfrac{1}{2}x -8$

⇒2y=−x-16

⇒x+2y+16=0

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We know that, The coordinates of point of contact is given by

$\bigg(\dfrac{a}{m^2} \ , \, \dfrac{2a}{m}\bigg)$

$\text{Point of contact = } \mathrm {\bigg(\dfrac{4}{(\frac{1}{4}) }, \dfrac{8}{(-\frac{1}{2} )} \bigg) } = (16,-16)$

Hope it helps!!

Answer 2

Step-by-step explanation:

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