Question

A hyperbola passes through the point \(P\left(\sqrt{2},\sqrt{3}\right)\) and has foci at \((\pm2,0)\). Then the tangent to this hyperbola at P also passes through the point :

• A. \(\left(2\sqrt{2},3\sqrt{3}\right)\)
• B. \(\left(\sqrt{3},\sqrt{2}\right)\)
• C. \(\left(-\sqrt{2},-\sqrt{3}\right)\)
• D. \(\left(3\sqrt{2},2\sqrt{3}\right)\)

Hint:

Use \(a^2+b^2=c^2=4\) and the point P to find \(a^2\) and \(b^2\), then write the tangent \(\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1\).

Answer 1

The Correct Option is A

Let the hyperbola be \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) with \(ae=2\), so \(a^2+b^2=4\).
Passing through P: \(\frac{2}{a^2}-\frac{3}{b^2}=1\). With \(a^2=A\): \(\frac{2}{A}-\frac{3}{4-A}=1\Rightarrow 8-5A=4A-A^2\Rightarrow A^2-9A+8=0\Rightarrow A=1\) or \(8\). Since \(A<4\), \(a^2=1\), \(b^2=3\).
Hyperbola: \(x^2-\frac{y^2}{3}=1\). Tangent at P: \(\sqrt{2}\,x-\frac{\sqrt{3}\,y}{3}=1\), i.e. \(\sqrt{2}\,x-\frac{y}{\sqrt{3}}=1\).
Check \(\left(2\sqrt{2},3\sqrt{3}\right)\): \(4-3=1\), so it lies on the tangent. (For \((\sqrt{3},\sqrt{2})\): \(\sqrt{6}-\frac{\sqrt{2}}{\sqrt{3}}\ne1\); options 3 and 4 also fail.)
Answer: (1) \(\left(2\sqrt{2},3\sqrt{3}\right)\)