Question

The locus of the point of interaction of two tangents of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ which make an angle α with one another is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let P(x₁, y₁) be a point in the locus.
Equation of a tangent to the hyperbola is $y=mx\pm \sqrt{a^2{m}^2-b^2}$
If this tangent passes through P then$y_1=m{x}_1\pm \sqrt{a^2{m}^2-b^2}\Rightarrow y_1-m{x}_1=\pm \sqrt{a^2{m}^2-b^2}\Rightarrow (y_1-m{x}_1{)}^2=a^2{m}^2-b^2\Rightarrow y_1^2+m^2{x}_1^2-2x_1{y}_{1}m=a^2{m}^2-b^2\Rightarrow (x_1^2-a^2)m^2-2x_1{y}_{1}m+(y_1^2+b^2)=0\to (1)$
If $m_1,m_2$ are the slopes of the tangents drawn from P to the hyperbola, then $m_1,m_2$
become the roots of (1).
$\therefore m_1+m_2=\frac{2x_1{y}_1}{x_1^2-a^2},m_1{m}_2=\frac{y_1^2+b^2}{x_1^2-a^2}$
$Nowtan\alpha =\frac{m_1-m_2}{1+m_1{m}_2}\Rightarrow (1+m_1{m}_2{)}^{2}ta{n}^2\alpha =(m_1-m_2{)}^2\Rightarrow (1+m_1{m}_2{)}^{2}ta{n}^2\alpha =(m_1+m_2{)}^2-4m_1{m}_2$
$\Rightarrow [1+\frac{y_1^2+b^2}{x_1^2-a^2}]ta{n}^2\alpha =\frac{4x_1^2{y}_1^2}{(x_1^2-a^2{)}^2}-\frac{4y_1^2+b^2}{x_1^2-a^2}\Rightarrow (x_1^2+y_1^2-a^2+b^2{)}^{2}ta{n}^2\alpha =(4x_1^2{y}_1^2-4(x_1^2-a^2)(y_1^2+b^2)\Rightarrow (x_1^2+y_1^2-a^2+b^2)ta{n}^2\alpha =4a^2{y}_1^2-4b^2{x}_1^2+4a^2{b}^2$
$\therefore$ The equation to the locus of P is $(x^2+y^2-a^2+b^2{)}^2=(4a^2{y}^2-4b^2{x}^2+4a^2{b}^2)co{t}^2\alpha$