Question

If tangents are drawn to the parabola $(x-3{)}^2+(y+4{)}^2=\frac{(3x-4y-6{)}^2}{25}$ at the extremities of the chord $2x-3y-18=0$, then angle between 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
$(x-3{)}^2+(y+4{)}^2=\frac{(3x-4y-6{)}^2}{25}$
The 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 with the standard parabola,we get
$\text{Focus}S\equiv (\alpha ,\beta )\equiv (3,-4)$

As $(3,-4)$ satisfies $2x-3y-18=0$
So, the given chord is focal chord

Tangents drawn at the extremities of focal chord are perpendicular and intersect at directrix so the angle is ${90}^{\circ }$