Question

Find the equation of the lines, that is parallel to the line 9x-8y=1, and tangent to the elipses 9x^2+16y^2=52.

Answer 1

Answer:

The equations of the lines that satisfy these conditions are

y = (9/8)x + (13/4)

And

y = (9/8)x - (13/4)

They can be written further as

8y = 9x + 26

and

8y = 9x - 26

Step-by-step explanation:

The line parallel to the line 9x - 8y = 1 has the same slope as 9x - 8y = 1

Writing the equation of the line in the (y = mx + c) form

8y = 9x - 1

y = (9/8)x - 1

So, the line(s) we are looking for has/have slope(s) of (9/8).

Its equation is then

y = (9/8)x + k

This line is then said to be tangent to the ellipses

9x² + 16y² = 52

So, since the line is a tangent to the ellipses, it will have the same coordinates as the ellipses at the point of contact.

9x² + 16y² = 52

y = (9/8)x + k

y² = (81/64)x² + (18k/8)x + k²

y² = (81/64)x² + (9k/4)x + k²

Substituting this into the ellipses equation,

9x² + 16y² = 52

9x² + 16[(81/64)x² + (9k/4)x + k²] = 52

9x² + (81/4)x² + 36kx + 16k² = 52

29.25x² + 36kx + 16k² - 52 = 0

Taking note that the two lines that will be tangent to the ellipses and parallel to the line given, to establish tangency, we make the discriminant of this equation 0. Hence, the roots of that equation will have equal roots.

So, the discriminant, b² - 4ac = 0

29.25x² + 36kx + 16k² - 52 = 0

a = 29.25

b = 36k

c = 16k² - 52

(36k)² - (4)×(29.25)×(16k² - 52) = 0

1296k² - 1872k² + 6084 = 0

576k² = 6084

k = ±(13/4)

So, the equation(s) of the line(s) required is

y = (9/8)x ± (13/4)

So, the equations include

y = (9/8)x + (13/4)

And

y = (9/8)x - (13/4)

Multiplying by 8, we obtain

8y = 9x + 26

8y = 9x - 26

Hope this Helps!!!