Question

If tangents are drawn to the parabola $(x-2{)}^2+(y-3{)}^2=\frac{(3x+4y-5{)}^2}{25}$ at the extremities of the chord $3x-y-3=0$, then angle between the tangents is

• A. ${45}^{\circ }$
• B. ${90}^{\circ }$
• C. ${60}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is B ${90}^{\circ }$

Given parabola is
$(x-2{)}^2+(y-3{)}^2=\frac{(3x+4y-5{)}^2}{25}$

General equation of parabola whose focus is $(\alpha ,\beta )$ and directix equation $ax+by+c=0$ is
$(x-\alpha {)}^2+(y-\beta {)}^2=\frac{(ax+by+c{)}^2}{a^2+b^2}$
Comparing the given parabola, we get
$\text{Focus}S\equiv (\alpha ,\beta )\equiv (2,3)$
Equation of directrix as
$3x+4y-5=0$

Given line is $3x-y-3=0$, which passes through $(2,3)$
So, the given chord is a focal chord

Therefore, the tangents drawn at the extremities of focal chord are perpendicular.