JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Ellipse

3 x + 4 y = 122

is a tangent to the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1

for some

a \in R

, then the distance between the foci of the ellipse is:

Answer: (B)

Solution

Write the tangent as

y = - \frac{3}{4} x + 32

, so its slope is

m = - \frac{3}{4}

and its intercept

c = 32

. The tangency condition

c^{2} = a^{2} m^{2} + b^{2}

with

b^{2} = 9

gives

18 = \frac{9 a ^{2}}{16} + 9

, so

a^{2} = 16

. With the major axis along the

x

-axis, the eccentricity is

e = 1 - \frac{9}{16} = \frac{7}{4}

, and the distance between the foci is

2 a e = 2 (4) \frac{7}{4} = 27

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option