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

Use the standard formula for the angle between tangents from \((x_1,y_1)\) to \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\): \(\tan\theta=\frac{2ab\sqrt{\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1}}{x_1^2+y_1^2-a^2-b^2}\) (absolute value).
Here \(a^2=\frac53,\ b^2=\frac52,\ ab=\frac{5}{\sqrt6}\), \((x_1,y_1)=(1,p)\). Then \(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1=\frac{3+2p^2-5}{5}=\frac{2p^2-2}{5}\) and \(x_1^2+y_1^2-a^2-b^2=p^2-\frac{19}{6}\).
Try the natural number \(p=2\): the numerator is \(\frac{10}{\sqrt6}\sqrt{\frac65}=\frac{10}{\sqrt5}\) and the denominator is \(\frac56\), giving \(\tan\theta=\frac{10}{\sqrt5}\cdot\frac65=\frac{12}{\sqrt5}\). It matches; \(p=1\) gives zero angle and larger \(p\) gives smaller values, so \(p=2\).
The parabola \(y=x^2-2x+3\) has vertex at \(x=-\frac{b}{2a}=1\), \(y=2\), i.e. \((1,2)\).
Distance to \((-7,8)\) \(=\sqrt{8^2+6^2}=10\).
Answer: 10