JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Ellipse

Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at

(3, - 4)

, one focus at

(4, - 4)

and one vertex at

(5, - 4)

. If

m x - y = 4, m > 0

is a tangent to the ellipse

E

, then the value of

5 m^{2}

is equal to ______

Answer: 3

Solution

The centre

(3, - 4)

, the focus

(4, - 4)

and the vertex

(5, - 4)

all lie on the line

y = - 4

, so the major axis is horizontal with

a = 5 - 3 = 2

and

c = 4 - 3 = 1

, giving

b^{2} = a^{2} - c^{2} = 3

. The ellipse is

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 4 ) ^{2}}{3} = 1

Writing

X = x - 3

and

Y = y + 4

, the line

m x - y = 4

becomes

m (X + 3) - (Y - 4) = 4

, that is

Y = m X + 3 m

. A line

Y = m X + λ

touches

\frac{X ^{2}}{4} + \frac{Y ^{2}}{3} = 1

when

λ^{2} = 4 m^{2} + 3

, so

(3 m)^{2} = 4 m^{2} + 3 ⟹ 5 m^{2} = 3

Hence

5 m^{2} = 3

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer