JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Parabola

If two tangents drawn from a point

(α, β)

lying on the ellipse

25 x^{2} + 4 y^{2} = 1

to the parabola

y^{2} = 4 x

are such that the slope of one tangent is four times the other, then the value of

(10 α + 5)^{2} + (16 β^{2} + 50)^{2}

equals ______.

Answer: 2929

Solution

Parametrize the ellipse:

α = \frac{1}{5} cos θ

β = \frac{1}{2} sin θ

.

Tangent to

y^{2} = 4 x

y = m x + \frac{1}{m}

. Passing through

(α, β)

\frac{1}{2} sin θ = m \cdot \frac{1}{5} cos θ + \frac{1}{m} ⟹ m^{2} \frac{cos θ}{5} - m \frac{sin θ}{2} + 1 = 0.

Let the slopes be

m_{1} = 4 m_{2}

. Then

m_{1} + m_{2} = \frac{5 s i n θ}{2 c o s θ}

and

m_{1} m_{2} = \frac{5}{c o s θ}

.

Substituting

m_{1} = 4 m_{2}

5 m_{2} = \frac{5 sin θ}{2 cos θ}, 4 m_{2}^{2} = \frac{5}{cos θ} .

Eliminating

m_{2}

cos θ = \frac{- 5 \pm 29}{2}

.

Thus

10 α + 5 = \pm 29

and

16 β^{2} + 50 = \pm 1029

(10 α + 5)^{2} + (16 β^{2} + 50)^{2} = 29 + (1029)^{2} = 29 + 2900 = 2929.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer