Hyperbola
Tangents from External Point — PQ² / (αβ)
nta_pyq_2023_apr
Grade 11
Question:
Let $m_1$ and $m_2$ be the slopes of the tangents drawn from the point $P(4,1)$ to the hyperbola $H:\ \dfrac{y^2}{25}-\dfrac{x^2}{16}=1$. If $Q$ is the point from which tangents with slopes $|m_1|$ and $|m_2|$ make positive $x$-intercepts $\alpha$ and $\beta$, then $\dfrac{(PQ)^2}{\alpha\beta}$ is equal to
Step-by-Step Solution
Key Concept: Tangent to $H$: $y=mx\pm\sqrt{25-16m^2}$. Passing through $(4,1)$: $4m^2-m-3=0\Rightarrow m_1=1,\ m_2=-\frac{3}{4}$.
$Q=(-4,-7)$. $(PQ)^2=128$. $\alpha\beta=16$. $\frac{128}{16}=8$.
Correct Answer: 8