Question

Two tangents are drawn to the curve $4x^2-9y^2=36$ such that the product of their slopes is $4$, then the locus of their point of intersection is

• A. is an ellipse with $e^2=\frac{4}{5}$
• B. is an ellipse with $e^2=\frac{3}{5}$
• C. is an hyperbola with $e^2=\frac{5}{4}$
• D. is an hyperbola with $e^2=5$

Answer 1

The correct option is D is an hyperbola with $e^2=5$

Equation of the tangents
$4x^2-9y^2=36\Rightarrow \frac{x^2}{9}-\frac{y^2}{4}=1\Rightarrow a=3,b=2$
Equation of the tangents
$y=mx\pm \sqrt{(am{)}^2-b^2}$
Let the tangents passes through $(h,k)$ then
$m^2(h^2-a^2)-2hkm+k^2+b^2=0m_1{m}_2=\frac{k^2+b^2}{h^2-a^2}=4k^2+b^2=4(h^2-a^2)$
So the locus of the point
$\Rightarrow 4x^2-y^2=b^2+4a^2\Rightarrow 4x^2-y^2=40\Rightarrow \frac{x^2}{10}-\frac{y^2}{40}=1\Rightarrow e^2=1+\frac{40}{10}=5$