Question

Tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ make angles ${\theta }_1$ and ${\theta }_2$ with the major of the ellipse such that $\tan ({\theta }_1+{\theta }_2)=k$ constant. The locus of the point of intersection of the tangents $(y^2-b^2)=$

• A. $\frac{k}{2}xy$
• B. $\frac{2}{k}xy$
• C. $\frac{xy}{k}$
• D. $kxy$

Answer 1

The correct option is B $\frac{2}{k}xy$

Let the point of intersection of tangent be (h,k)

Then equation of tangent with slope m is given by $y=mx+\sqrt{a^2{m}^2+b^2}$

(h,k) lies on the tangent

then $k=mh+\sqrt{a^2{m}^2+b^2}$

$\to (a-h{)}^2{m}^2+2hkm+b^2-k^2=0$

we get $m_1+m_2=\frac{-2hk}{a^2-h^2}$

given that

$tan{\theta }_1+tan{\theta }_2=k$

$\to m_1+m_2=k\to \frac{-2hk}{b^2-k^2}=t$

therefore,

$2xy=k(y^2-b^2)$

$(y^2-b^2)=\frac{2xy}{k}$