Let m_1 and m_2 be the slopes of the tangents drawn from the point P(4,1) to the hyperbola…

Question

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 $\alpha $ and $\beta $ on the $x$-axis, then $\frac{(PQ{)}^{2}}{\alpha \beta }$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 8

Accepted Answer

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 :

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

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

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

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

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

which simplifies to the quadratic equation :

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

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,

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

Solutions · 1