JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Hyperbola

Let

m_{1}

and

m_{2}

be the slopes of the tangents drawn from the point

P (4, 1)

to the hyperbola

H : \frac{y ^{2}}{25} - \frac{x ^{2}}{16} = 1

. If

Q

is the point from which the tangents drawn to

H

have slopes

∣ m_{1} ∣

and

∣ m_{2} ∣

and they make positive intercepts

α

and

β

on the x axis, then

\frac{( P Q ) ^{2}}{α β}

is equal to _____.

Answer: 8

Solution

The tangent to the hyperbola

\frac{y ^{2}}{25} - \frac{x ^{2}}{16} = 1

y = m x \pm 25 - 16 m^{2}

.

Passing through

P (4, 1)

1 = 4 m \pm 25 - 16 m^{2}

. Squaring:

1 - 8 m + 16 m^{2} = 25 - 16 m^{2}

, giving

32 m^{2} - 8 m - 24 = 0

4 m^{2} - m - 3 = 0

.

Solving:

m = 1

m = - \frac{3}{4}

, so

∣ m_{1} ∣ = 1

and

∣ m_{2} ∣ = \frac{3}{4}

.

Tangents with positive slopes

1

and

\frac{3}{4}

making positive x-intercepts:

Intersection point:

Q (- 4, - 7)

. Distance:

P Q^{2} = (4 - (- 4))^{2} + (1 - (- 7))^{2} = 64 + 64 = 128

.

Therefore:

\frac{P Q ^{2}}{α β} = \frac{128}{\frac{16}{3} \times 3} = \frac{128}{16} = 8

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer