JEE Main 2023 — Hyperbola Question with Solution

Equation of tangent to the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

$$y=mx\pm \sqrt{a^2-b^2{m}^2}$$

Given the hyperbola $H:\frac{y^2}{25}-\frac{x^2}{16}=1$, the equation of the tangent to this hyperbola can be written as :

$$y=mx\pm \sqrt{25-16m^2}$$

We know that the tangents pass through the point $P(4,1)$, which gives us the equation :

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

Squaring both sides to get rid of the square root, we obtain :

$$(4m-1{)}^2=25-16m^2$$

which simplifies to the quadratic equation :

$$4m^2-m-3=0.$$

Solving this equation, we find the roots $m_1=1$ and $m_2=-\frac{3}{4}$, which are the slopes of the tangents.

Given that we are interested in the positive values of the slopes, we consider $|m_1|=1$ and $|m_2|=\frac{3}{4}$.

The equations of the tangents are then :

1) $y=x-3$ and

2) $y=\frac{3}{4}x-4$.

The x-intercepts of these lines (when $y=0$) are given by $x=\alpha =3$ for the first line and $x=\beta =\frac{16}{3}$ for the second line.

The intersection point $Q$ of these tangents is found by solving the system of equations $y=x-3$ and $y=\frac{3}{4}x-4$, which gives $Q(-4,-7)$.

The square of the distance $PQ$ is then $(-4-4{)}^2+(-7-1{)}^2=128$.

Therefore, $$\frac{(PQ{)}^2}{\alpha \beta }=\frac{128}{3\times \frac{16}{3}}=8$$