Question

Line $2x+y=3$ cuts the ellipse $4x^2+y^2=5$ at $P$ and $Q$. If $\theta$ is the angle between normals at $P$ and $Q$, $\tan \theta =$

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{3}{5}$
• D. $5$

Answer 1

The correct option is C $\frac{3}{5}$

Solving $y=3-2x$ and $4x^2+y^2=5$,

we get points of intersection as point $P(\frac{1}{2},2)$ and $Q(1,1)$.

The equation of tangents at these points are

$4.x.\frac{1}{2}+2.y=5\Rightarrow m_1=-1$ and $4.x.1+y.1=5\Rightarrow m_2=-4$

$\therefore$ Slope of respective normals are

${m^{\prime }}_1=1$ and ${m^{\prime }}_2=\frac{1}{4}$

If $\theta$ is the angle between them, then

$\tan \theta =\left|\frac{{m^{\prime }}_1-{m^{\prime }}_2}{1+{m^{\prime }}_1{m^{\prime }}_2}\right|=\frac{1-\frac{1}{4}}{1+1.\frac{1}{4}}=\frac{3}{5}$