Question

For \(p \in \mathbb{N}\), if the angle between pair of tangents drawn to the ellipse \(3x^{2}+2y^{2}=5\) from the point \((1, p)\) is \(\tan^{-1}\left(\frac{12}{\sqrt{5}}\right)\), then the distance of the vertex of the parabola \(y=x^{2}-px+p+1\) from the point \((-7,8)\) is equal to ________.

Hint:

Find the slopes of the two tangents from \((1,p)\) using \(c^2=Am^2+B\), then use \(\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\) to get \(p\).

Answer 1

Correct Answer: 10

The ellipse is \(\frac{x^2}{5/3}+\frac{y^2}{5/2}=1\). A tangent \(y=mx+c\) satisfies \(c^2=\frac53m^2+\frac52\). The tangent through \((1,p)\) is \(y=mx+(p-m)\), so \((p-m)^2=\frac53m^2+\frac52\).
Simplifying: \(4m^2+12pm-6p^2+15=0\). Hence \(m_1+m_2=-3p\), \(m_1m_2=\frac{15-6p^2}{4}\).
\((m_1-m_2)^2=9p^2-(15-6p^2)=15(p^2-1)\) and \(1+m_1m_2=\frac{19-6p^2}{4}\).
\(\tan\theta=\frac{4\sqrt{15(p^2-1)}}{|19-6p^2|}\). For \(p=2\): \(\frac{4\sqrt{45}}{5}=\frac{12\sqrt5}{5}=\frac{12}{\sqrt5}\), which matches. (\(p=1\) gives angle 0; for \(p\ge3\) the value falls well below \(\frac{12}{\sqrt5}\).) So \(p=2\).
Parabola: \(y=x^2-2x+3=(x-1)^2+2\), vertex \((1,2)\).
Distance from \((-7,8)\): \(\sqrt{(1+7)^2+(2-8)^2}=\sqrt{64+36}=10\).
Answer: 10